DOCUMENT RESUME
ED 085 548 CE 000 728
AUTHOR Pierro, Mike; And OthersTITLE Geometry: Career Related Units. Teacher's Edition.INSTITUTION Minnesota State Dept. of Education, St. Paul. Div. of
Vocational and Technical Education.; RobbinsdaleIndependent School District 281, Minn.
PUB DATE Jan 73NOTE 270p.
EDRS PRICEDESCRIPTORS
MF-$0.65 HC-$9.87*Career Education; *Curriculum Enrichment; *Geometry;*High School Students; Resource Units; Senior HighSchools; *Unit Plan
ABSTRACT/ Using six geometry units as resource units, the
document explores 22 math-related careers. The authors intend thedocument t9/provide senior high school students with careerorientation and exploration experiences while they learn geometryskills:-Tbe units are to be considered as a part of a geometrycourse, not a course by themselves. The six geometry units (righttriangles and the Pythogorean theorem, polygons and their areas,parallel lines, standard constructions, volume, and circlerelationships) may either be studied first or used as resource unitsas the student works in any of the career units: printing and thegraphic arts, heavy equipment operator, fashion and apparel design,navigation, painting and paperhanging, landscape technology,carpenter, architecture and drafting, optical technician, sheetmetal, engineering, machinist, cement workers, forestry, electrician,general contractor, home planning, cabinetmaking, plumbing and pipefitting, surveyor, outdoor advertising, and space. The teacher'sedition contains an answer key. (AG)
FILMED FROM BEST AVAILABLE COPY
11
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CAREER RELATED UN TS
MINNESOTA STATE DEPARTMENTOF EDUCATION
VOCATIONAL TECHNICAL DIVISION
CAPITOL SQUARE BUILDING`SAINT PAUL, MINNESOTA
U.S Ci.1PARTMENT OF HEALTH.EDUCATION& WELFARENATIONAL INSTITUTE OF
EDUCATIONTHIS DOCUMENT HAS BEEN REPRODUCEO EXACTLY AS RECEIVED FROMTHE PERSON OP ORGANIZATION ORIGINAT ING IT POINTS OF VIEW OR OPINIONSSTATEO 00 NOT NECESSARILY REPPESENT OFFICIAL NATIONAL INSTITUTE OFEDUCATION POSITION OR POLICY
State of Minnesota
DEAR EDUCATOR:
Department of EducationCapitol Square, 550 Cedar StreetSt. Paul, Minnesota 55101
THIS DOCUMENT IS PART OF A PROJECT FUNDED BY THE MINNESOTA
DEPARTMENT OF EDUCATION, DIVISION OF VOCATIONAL-TECHNICAL
EDUCATION IT WAS DEVELOPED TO SHOW HOW GEOMETRY CURRICULUM
CAN BE ENHANCED BY ADDING REAL-LIFE CAREER ORIENTED AREAS
REQUIRING MATH SKILLS.
IT SHOULD PROVIDE STUDENTS WITH CAREER ORIENTATION AND EXPLORA-
TION EXPERIENCES AS WELL AS BEING AN EFFICIENT, RELEVANT TEACHING
VEHICLE FORHELPING STUDENTS ENJOY LEARNING GEOMETRY SKILLS.
c_
Robert M. Madson, DirectorProgram OperationsDIVISION OF VOCATIONAL-TECHNICAL EDUCATION
RMM:LBK/sns
I. GE
OM
ET
RY
RE
SO
UR
CE
UN
ITS
-- TA
BLE
OF
CO
NT
EN
TS
I.
Right.Triangles and the PythagoreanTheorem
2..
Polygons .and Their Areas
3.
Parallel Lines
4.
Standard Constructions
-5.
Volume
6.
Circle Relationships
GEOMETRY CAREER RELATED UNITS
Teacher's Edition
A Joint Project between the Minnesota State Department ofVocational Education and the Robbinsdale Area Schools
Writers
Mike Pierro, Math Teacher, Robbinsdale Senior High SchoolClaude Paradis, Math Teacher, Cooper Senior High SchoolDoug Svihel, Math Teacher, Armstrong Senior High SchoolArnie Grangaard, Vocational Counselor, Armstrong Senior High School
January, 1973
Cliff Helling William HockVocational Coordinator Math Consultant
Independent School District 281Robbinsdale Area Schools4148 Winnetka Avenue North
Minneapolis, Minnesota 55427
TABLE OF CONTENTS
Introduction 1
Procedure 1
List of Units 3
Student Record Sheet 4
Appendix: Answer Keys 7
INTRODUCTION
This project is an attempt to bring a practical approach to a senior high
school geometry course. At the onset of the project, this seemed like a
simple matter. The writing team began their work by investigating many
careers. They found many isolated cases where geometry is used, but not
in sufficient quantity to build a unit. This forced the team, in some
cases, tc work with broader categories of career's'.
The units produced were written using as practical an approach as possible,
and only after interviewing a person actually involved in that jcb. The
purpose of the interview was to find the geometry-related tasks of each
career.
The total package turned out to be a set of consumable units written in
narrative form including six geometry resource units and 22 career units.
The units are intended to be used by the students, written in, and kept.
Note: These units are meant to be used as a Rart of a geometry course and
are not meant to be used exclusively by themselves.
PROCEDURE
There are several procedures that could be followed in using these materials.
One would be to work through all of the resource units with the class and
then have students as individuals select the career units they are interested
in. A second way would be to allow the students as individuals to begin
work in the career units and then refer to the resource units as they
are needed.
Evaluation: A most successful way of dealing with units of this type is
to assign each unit a point value. A student then must earn a certain
1
minimum number of points working through units of his choice.
CAREER-RELATED GEOMETRY
Answer Keys: The answer keys to these units should be made available
to the student by the units, not in book form. This keeps the student
from checking the answers for one unit and copying the answers to his
next unit.
Resource Materials: Resource materials of value would be:
Occupational Outlook Handbook
Encyclopedia of Careers, Volumes 1 and 2
LIST OF UNITS
I. Geometry resource units
1. Right Triangles and 'he Pythagorean Theorem2. Polygons and Their Areas3. Parallel Lines4. Standard Constructions5. Volume6. Circle Relationships
II. Career units
1. Printing and Graphic Arts2. Heavy Equipment Operator3. Fashion and Apparel Design4. Navigation5. Painting and Paperhanging6. Landscape Technology7. Carpenter8. Architecture and Drafting9. Optical Technician
10. Sheet Metal11. Engineering12. Machinist13. Cement Worker14. Forestry15. Electrician16. General Contractor17. Hcme Planning18. Cabinetmaking19. Plumbing and Pipe Fitting20. Surveyor21. Outdoor Advertising22. Space
3
I. Geometry Resource Units
Unit
STUDENT RECORD SHEET
Date Date AdditionalStarted Finishe Math Work
Related
Name
1. Right Trianglesand the
Pythagorean Theorem
2. Polygons andTheir Areas
3. Parallel Lines
4. StandardConstrictions
5. Volume
6. CircleRelationships
II. Career Units
Unit
STUDENT RECORD SHEET
Date Date AdditionalStarted Finished Tork
RelatedPry'
Name
1. Printing
.._ _
2. Heavy EquipmentOperator
3. Fashion and Apparel______csn
4. Navigator
5. Painting andWall Papering
,. Landscape Technology
7. Carpenter
8. Architecture andDrafting
9. Optical Technician
LO. Sheet Metal
Li. Engineering
L2. Machinist
3. Cement Worker
4. Forestry
3. Electrician
General Contractor
5
II. Career Units
STUDENT RECORD SHEET
Date Date Additional Relatedd Math Work
Name
17. Home Planning
18. Cabinetmaking
19. Plumbing andPipe Fitting
20. Surveyor
21. Outdoor Advertising
9.2. Space
6
RIGHT TRIANGLES AND THE PYTHAGOREAN THEOREM--ANSWER Kr
1. a. 90°
b. Right triangle
c. Yes 92
+ 122
= 152
Si + 144 = 225
a. 44C
b. AB or C
c. Right; Hypotenuse
3. a. No
b. Yes
4. a. 20
b. 10
c. 10 rr
5. a, c = 7 (17
b. x = 25 25 vIror2
POLYGONS AND THEIR AREAS -- ANSWER KEY
Type ofPolyaon
Sites A'sFormed
Degrees inEachL Total Degrees in
the Polygon
triangle 3 1. 180o 0
180
quadrilateral 4 2 180° 360°
pentagon 5 3 180o
540°
hexagon 6 40
180 720°
octagon 8 6 °1 80 1030°
decagon 10 8 180° 1440°
n-gon N N -2 180°180 (N-2).(180)
1. 900°
2. a. 108°b. 135°
3. 450
square
4. Area = 15.2 square
6. 369.5 square inches
7. 32 square inches
8. 320 square inches
inches
10
5. Area = 52.5 square inches
PARALLEL LINES -- ANSWER KEY
1. a. 50°
b. 130°
c. 50°
0d. 130
e. 50°
f. 130°
g. 50°
h. 130°
2. a. Z.6 or L4
b . Z 4 or 2=6
c. Equal
3. a; b. 2: 2 and 2.6
c; d. 4.3 and L.7
e. Equal
4. 4:3 and L6
5. Yes
6. 6
7.
/\f`
04 r
\
t).sr
Construct parallel lines as shown.To divide AB draw another Line similar to AG and join its endpoint M with B.Then construct lines.
11
3.
4.
STANDARD CONSTRUCTIONSANSWER KEY
1. Construct_L angle.2. Construct hypotenuse = 2 x short leg3. - 300-600-9004. Construct 2 4-bisectors for
inscribed center5. Drop 1 for radius length.6. Draw semicircle7. Construct 2 biscetors of sides
for circumscribed center8. Draw circumscribed 0
1. Construct radius through given point.2. Construct to radius at the
given point.
13
VOLUME ANSWER KEY
IV. a. 106.6 cubic yards
b. 5000 cubic feet
c. 185.2 cubic yards
V. V=Bxh. 71r. 62 18
= 29 x 36 x 18 = 2036.57 cubic7 1 1
14
inches
1. a.
2. a.
3.
CIRCLE RELATIONSHIPS -- ANSWER REY
b.
b.
c.
c.
a.
d.
b. c.
e.
4. a. 45 7. 125 9. a. 33.49 10. a. 41.04
b. 90 8. a. 6.98 b. 83.73 b. 1.45
c. 45 b. 18.32 c. 50.24 c. 8.26
d. 30 c. 12.56 d. 89.32 d. 2.46
5. a. 80 d. 1,308.33 11. 10.08
b. 50
6. 37
15
PRINTING AND GRAPHIC ARTS -- AMIE-a KEY
1. Root -2, hypotenuse
2. Root-4, long
3. Root-3, printers
4. Golden
5. Root-2, hypotenuse
6. 9" x 6" -- printers
1277 x 3-1 -- hypotenuse
7. (As on page 4)
8.
3"
HEAVY EQUIPMENT OPERATOR -- ANSWER KEY
1. 56' x 48' = 2688 square feet
2. 21,504 cubic feet = 796.4 cubic yards
3. 208 feet
4. V = 208' x10'
x = 173.3 cubic feet12
5. 21,677.3 cubic feet or 802.86 cubic yards
6. A = 54' x 11' = 594 square feet
7. A = 44' x 18' = 792 square feet =
8. Area of circle = 1,017.4
Area of sector = 254.3 square feet
9. 792 - 508.6 = 283.4 -11 283 square feet
10. Total area of driveway = 2,971 square feet
11. 1485.5 cubic feet (V = 2,971 x .5)
12. 55 cubic feet
13. Hours = 5.5
Cost = $247.50
14. 495 cubic feet or 18.33 cubic yards
17
1. 16 incries
2. 9 inches
3. Drawkngs vary.
4. Examples vary.
5. Examples vary.
6. 1.618
7. Examples vary.
8. Examples vary.
9. Examples vary.
FASHION AND APPAREL DESIGN -- ANSWER KEY
L8
NAVIGATION -- ANSWER KEY
1. Latitude, 40° N; longitude, 70° W
Latitude,50° S; longitude, 50° E
3. Latitude, 80° N; longitude, 80° N
4, Latitude, 20° S; longitude, 30° E
5. Latitude, 80° S; longitude, 80° W
6. C and E
7. - 10. See below.
8
N
11. 90° T
12. 135° T
L3. 60 miles in one hour; 20°
14. a. 110 Tb. Resultant ground speed = 170
15. Approximate 220 mph
16. 90 miles in one hour, approximately 25°
017. 115 T
s7
19
I
PAINTING AND PAPERHANGING -- ANSWER KEY
1. 707 square feet
2. Approximately 31 gallons = 3.535 gallons2
3. a. 32 17440 = 1216 square feet
b. 2.13 gallons
4. a. 38.016 coats (Use = Ir.)7
b. $190.08
5. $59.76
6. a. 514a square feet3
b. 831 square feet3
c. 5.528 rolls = 6 rolls
d. At $6.95 roll = $41.70
7. 12.365 gallons
height of gable end is4 2
= a feet
8. 625.04 square feet
9. 3.6 gallons
110. 53-
3square feet
LANDSCAPE TECHNOLOGY"-- ANSWER KEY
1. a. 247 nitrogen, 16% phosphorus, 12% potash
b. 27,000 squre feet
c. 151.87
2. 12.5 cups
3. a. 8.07 tons
b. 254 feet
c. 130.68 cubic yards
d. 35.64 pounds
4. a. Construct angle bisectors for bac angles.
b. Construct the J. bisector of each angle bisector.
5. $1,248.62
6. 310.86 square feet
7. 207.84 square feet
8. 117.15 cubic feet or 4.34 cubic yards
21
CARPENTER -- ANSWER KEY
1. sp.s' and 60.8'
2. 5I
3. 19'
4. 2p' approximate
5. 21' 6"
6. 4p8 square feet 15 sheets of 4' by 8' plywood
7. square feet
8. 17 and 12 approximate
2141
9. 6 squares
10. 1po cubic yard
11. 804 cubic feet.
1. Sketch
ARCHITECTURE AND DRAFTING -- ANSWER KEY
Draw AB 8 inches long.Divide AB with parallel lines into parts of ratio 2 4 5.
11 11 11At point t, with arc of radius AD, swing an arc about C.At point E, with arc of radius EB, swing an arc about E.Intersection of these arcs call M.Required triangle is MDE.
2. Hints: 1. Sum of acute angles is 90°2. Hypotenuse of right triangle is the diameter of the circle
that circumscribes it.
3. Follow directions in problem carefully.
A
23
ARCHITECTURE AND DRAFTING -- ANSWER KEY
4.
I II
III
1
_
4-- 1
3%
Ski"Top
Side
The above are sketches of elevations. Plan should be laid out as shownon page 5 at bottom.
5. Approximately 5.25 or S inches.4
6. a. Approximately 3.50b. Approximately 0.0847 inches
7. 25" x 162"8
8. 7 inches
9. Plan should be 5" by- 4-1"16
24
1. 350
2. 200
3. Yes0
corresponding L 's formed by parallel lines are 1;r
OPTICAL TECHNICIAN -- ANSWER KEY
A
MEMO IMES
4. 60°
5. Inward
6. A..
Normal B
7. --"'
C
8. a. RTBA
b. 07BA
c. ORA
d. it TB
e. PORT
9. ORT:0
10. ORA or a.TB
90oAngle of reflection
SHEET METAL -- ANSWER KEY
1. a. 31.4 square feet (Use = 3.14.)
b. 4 sheets
2. a. d = 2 408.7 inches = 40.4 inches
b. A = 81.25 square feet
3. a. 70.65 cubic feet
b. = w = h = 4.14 feet
4. 555 square feet = 69.375 square feet8
5. 25° and 65°
6. 4.71 inches
7. The length of one side
8. Students' diagram should show dividing the circle into six equal parts. Construct
this by walking the compass (opened to length of circle radius) around the
circumference. Now find centers of holes to be drilled.
9.
26
1. 22.11 souare feet
2. a.
3. 2781.84 pounds
ENGINEERING -- ANSWER KEY
b. 195 cubic yards
4. 2563.22 square rods = 77,537.405 square yards
5. b. Area = 10,000 square yards
For the map, follow the hint on page 5.
6. 20 Vrafeet = 28.20 feet
7. a. Approximately 620 feet
b. They are perpendicular
8. 868.05 cubic feet
9. a. 8 feet
b. 4 feet
c. 12.8 feet
d. 10.8 feet
e. 32.3 feet
27
1. 22.11 square feet
2. a.
3. 2781.84 pounds
ENGINEERING -- ANSWER KEY
5b. 195-9
cubic yards
4. 2563.22 square rods = 77,537.405 square yards
5. b. Area = 10,000 square yards
For the map, follow the hint on page 5.
6. 20 V-' feet = 28.20 feet
7. a. Approximately 620 feet
b. They are perpendicular ( OT ,L TP).
8. 868.05 cubic feet
9. a. 8 feet
b. 4 feet
c. 12.8 feet
d. 10.8 feet
e. 32.3 feet
27
MACHINIST -- ANSWER KEY
1. 60
2. + .707
3. 4r8--
4. a. .8660
b. 2.3"
c.
5.
Step 1. 45°
1°Step 2. 227
Step 3. Sine 221° 4.-- .38262
Step 4. a 1.913
28
Step 5. 2a 3.826
Step 6. Sine 45 = '-AO
.7071 =
.7071A = 1
A =1
= 1.414.7071
CEMENT WORKER -- ANSWER KEY
ti
1. mLLA = 60
2. 8'
3. A = 2 bh (or A = \(s -a) (s-b) (s-c) ) = 2 x 8 x 6.928 = 27.7 square feet
4. Total area = 166.27
5. 166.27 x .333 = 55.4 or 55--1
cubic feet2
6. 27 cubic feet = 1 cubic yard 2.05 = 2 cubic yard
7. a. Total side area = 88 square inches + 176 square inches + 264 square inches+ 1248 square inches = 1776 square inches
1b. 12--
3square feet
1c. V = 12--
3square feet x,4 feet = 49,1 cubic feet
1d. 49--
3- 16
49.33 - 16.4 = 32.9 =- 33 cubic feet = 1.22 cubic yards
8. 5358 square feet V = 5358 x A- = 1784.2 cubic feet V = 66.08 cubic yards12
18'
30' 8
/9
29
9. Ratios of corresponding sides are in proportion
29
CEMENT WORKER -- ANSWER KEY
10. 1:3
11. 1:9 (smaller A's A = 24 square feet, larger A = 216)
12. 9 times
13. The areas vary as the square of the ratio of the sides
FORESTRY -- ANSWER KEY
b = 114 approximately
4. 20.73 feet
5. 53.98 inches
6. a. 90
b. the distances are equal
c. distance DC
7. a. 110°
b. 225°
0c. 310
32
1.B
2.
C
ELECTRICIAN -- ANSWER KEY
182= 15
2+ X
2
2 2 2
18 - 15 = X
324 - 225 = X2
299 = X
10 = X
C2= 100
2+ 144
2
2
C = 10,000 + 20736
2C = 30,736
144b
3. a. C = 3.14 x 7 = 21.98"0
b. m DC"
=360
0 of C115 .32 x 21.98 = 7.03"
c. Total from A to B = 8i.-" + 7.03" + 8" = 23.53"
GENERAL CONTRACTOR -- ANSWER KEY
1. $201.375 = $201.38
2. $270
3. a. 1338 square feet
b. 41 bundles
c. 42 bundles
d. $210
4. Approximately 10.88 cubic yards
5. 6.72 hours
$45.70
6. a. 54'-11"
b. 1771-10"
c. 14 + 1 = 15 joists
d. $90.60
7. Area = 2244 square rods
14 acres
Cost = $42,000
8. Gravel = 75 cubic yards
Sand = 561 cubic yards
Cement 182 cubic yards4
9. 62.8 cubic feet = 2.326 cubic yards
Cost = $48.85
10. 194 feet (UseTr=4 7
11. 606 square feet
34
C. a = 5 feet
D. P = 19 feet
E. S = 9.5
2.
C. a = 6 feet
D. P = 14 feet
E. S = 7
3.
HOME PLANNING -- ANSWER KEY
b = 5 feet c = 9 feet
A = 9.8 square feet
b = 2 feet c = 6 feet
A = 5.9 square feet
C. a = 5' b = 5' c = 9'
D. P = 19'
E. S = 9.5 A = 9.8 square feet
4. A. P = 17 feetB. A = approximately 12 square feet
5. A. P = 15B. A = approximately 10 square feet
6. A. P = 12B. A = 6 square feet
,Zr.
Srimple8
8.12S
10. ="8
11,
12.
13.
14.
5,1
8
2,1
8
51116
311
8
15. 15"8
16.
17.
18.
16
3I116
28
19. 78'
20. 18'
21. 1404 square feet
22. 30'-
23. 2826 square feet
24. 18'
25. 1017 square feet
3_6
KEY
26. 1809 square feet
27. 3213 square feet
28. 1070 cubic feet
29. 40 cubic yards
30. $840
31. $430
32. Driveway B.
1. e. A = 6"
B = llu4
C =4
D = 7"
E = 6"4
F =4
it
G =4
H = 231w
a. 40"
b. 20"
- 135l6720 o
2
4.
CABINETMAKING -- ANSWER KEY
Let R1
= 2
R2
= /
R31
2
37
CABINETMAKING -- ANSWER KEY
b. Let R1 be radius of CD
Use K as center and draw arc with radius R1 1
Construct line parallel to AB, 1" above AB
Where parallel line and arc of radius R1 intersect locates unkown
arc's center.
5. 169.56 inches
6. .86 square feet
7. Use a large sheet of paper or cardboard. Need 13" x 28" for end view.
13, 0-
--inches
4
9. The b16,!ctors of two Lhords intersect at the center of the circle.
38
PLUMBING AND PIPE FITTING -- ANSWER KEY
1. 36
2. 16
3. 20
4. Radius of single pipe =61.
16 3'81
Diameter of single pipe = 7.61
5. Examples will vary. Doubling the diameter will increase capacity four times.
6. Area of piston = !f 42 = 16'ir = 50.24 square inches
Total pressure = 50.24 x 80 = 4019.2
7. a. 45
b. 5
8. 3636 I=
9. 10C = 36
C =
39
SURVEYOR -- ANSWER REY
1. N 50° W
2. S 68° W
3. N 30° W
4. S 70° E
5. a. 15" x 6"
b. h = 25',
c. AB = 27',
CD = 39'
w = 20'
BC = 39'
DA = 36'
6.
a. 751-11"2
b. N 7° -10° W (Allow for some error)
7. From point A N 80° E 100'; N 90° E 150'; S 50°E 125'; S 0° 75'; N 90° W 200';
N 35° W 62.5'
8. 3" = 150'
9. N 60° W
40
1.
2.
OUTDOOR ADVERTISING -- ANSWER KEY
31 (3.428) 21. a. 1 X 21 (nearest 1)4 4
131" b. Poster2
3. 18" c. Bleed poster
4. 27" 22. a. 1 X 2-1 (nearest 1)
4 4
5. 36" b. Poster
6. 45" c. Bleed poster
7. 8'-11" 23. a. 1 X 541)
4 4
8. 19'-6" This is one of the obvious exceptionsto the main rule.
9.021002 (2.187)
b. Bulletin10. 91-7"
c. Embellished11. 21'-7"
(2.252)2115Z2
13. 10'
14. 22'
15.12-5
(2.200)
16. 672 square feet
17. 872 square feet
18. a. 1 X 3 (nearest 1)4
b. Bulletin
c. Embellished
19. a. 1 X 21 (nearst1)
4 4
b. Poster
c. 30-sheet poster
20. 1)a. 1 X 244 4
b. Poster
c. 30-sheet poster
41
SPACE -- ANSWER KEY
1. a. 1,040
b. 173 i3
c. 20.02 centimeters
2. a.VT17
b.21)7577r
3. a. vr(r+h)2
4.
5. 1,737,5 kilometers
6. 88.1 days
7.
or
/17572777.
b. 895.54
c. 889.94
d. 895.54
e. 889.94
f. .63%
a. 395.09
b. 388.84
c. 1.587
a. 1.23%
b. 2.03%
c. 3960 miles
d. 42.46%
e. 42.46%
42
RIGHT TRIANGLES
Many years ago Egyptians used the idea that a triangle with sides 3,4, and 5
units long is a right triangle. They used this fact in construction and sur-
veying. The Babylonians knew other right triangles besides the 3-4-5 triangle.
For example, they were aware of the fact that a triangle whose sides were 5,
12, and 13 units long is also a right triangle. vs,
Let's investigate a relation between the lengths of the sides of a right tri-
angle. Physical representations of square regions can be cut from graph paper
and arranged so that their edges outline a triangle as shown below.
I. II
Measure the angles of the triangles above. You should discover that all three
triangles have a right angle and therefore are right triangles.
RIGHT TRIANGLES
Consider the lengths
23- + 4- = 5
9 + 16 = 25
3, 4, and 5 of the sides of triangle I.
Consider the lengths
5-9+ 12- = 13
25 + 144 = 169
5, 12, and 13 of the sides of triangle II.
Consider the lengths 6, 8, and 10 of the sides of triangle III.
6-9
+ 82
= 102
36 + 64 = 100
You can see it is possible for certain squares having sides equal in length to
9 9 9a, h, and c and respective areas of a- , h- , and c- square units to be arr-
anged so as to outline a right triangle if a- + b- = c as shown with the
3-4-5, 5-12-13, and 6-8-10 triangles above.
1. Cut three squares out of a sheet of graph paper having lengths 9, 12,
and 15. Arrange them on a sheet of paper so that their edges form a
triangle.
a. What is the measure of the largest angle?
b. What kind of triangle is formed?
c. If a, b, and c represent the lengths of the sides of tha
triangle, is it true that a2 b2 = c2
Show this!
Consider once again the 3-4-5 triangle. We know now that 32+ 4
2= 5
2and
that the triangle is a right triangle. Look again at the geometric interpre-
tation of this fact on the following page.
2
Notice that the length of the side opposite angle A is represented by the
letter a . The length of the side opposite angle B is represented by b ,
and the length of the side opposite angle C by c.
2. a. Which angle ofQ ABC is a right angle?
b. BC and AC are called the "legs" of right triangle ABC. The
longest side of a right triangle is called the "hypotenuse."
What line segment is the hypotenuse of right triangle ABC?
c. The hypotenuse is the side opposite the
The right angle is always opposite the
angle.
The diagram on the following page shows that if each side of the right triangle
is one side of a square, then the area of the largest square is equal to the
sum of the areas of the two smaller squares.
3
RIGHT TRIANGLES
52 or.,25 square units
N4' or -t
16 square/s- / ! units.-/ b i 1
32 or_9 square
_. 1....11units
We call this relationship a2 + b2 = c 2 of the sides a, b, and c of any
right triangle the Pythagoream Theorem: that is,
"For any right triangle, the sum of the squares of the lengths of the legs is
equal to the square of the length of the hypotenuse."
3. The lengths of the sides of two triangles are given below. Are they
right triangles?
a. 9 in., 11 in., 12 in.
b. 12 ft., 16 ft., 20. ft.
The formula a2 + b2 = c2 allows us to find the length of one side of a
right triangle when you know the lengths of the other two sides. Suppose the
two legs were 33 inches and 44 inches. You could find the length of the hypotenuse c
in the following way.
a2 + b
2= c
2
332
+ 442
= c2
1089 + 1936 = c2
3025 = c2
77=4
RIGHT TRIANGLES
Remember you are looking for c and not Therefore, you must find the
"square root" of c2 to find c.
c =47727
c = 55 in.
A sfmple check on your answer is to multiply 55 times 55 and your product should
be 3025. Suppose you knew the length of one leg (any one) and the hypotenuse.
You could find the length of the other by first subtracting c2 - a 2 Or
C- h-9and then taking the square root. Going back to the previous problem,
if we know one leg is 33 inches and the hypotenuse is 55 inches, we can find the
length of the other leg like this:
If a2 + b2 = C2. , then b2 = c2
- a 9
55" - 332
2b = 3025 1089
b2
= 1936
b = 1936 = 44 in.
There are some special right triangles that lend some shortcuts in finding
lengths of sides. These are the 30° -60° - right triangle and the 45° -45° -
right triangle.
Consider the equilateral triangle below. Each angle is 60° . Now if we draw
an altitude as pictured, then two 30-60-right triangles are formed. The alti-
tude also bisects the side of the equilateral triangle that it is perpendicular
to.
5
RIGHT TRIANGLES
If the length of each side of the equilateral triangle is x, then each of the
two segments that the altitude divides the side of the triangle into are lx
as pictured on the previous page. Now we will find the length of the altitude
h by using a2 + b2 = c?.
h2 ( -x)2 = x22
'h2 = x2 lx2
4
h2 4 x2 1 x24 4
2 3 2h = x
4
3 x2
4
h _ x 13--2
h = 1 x 172
To summarize this, if you have a 30°-60°-right triangle, then
a. The length of the shortest leg is one-half the length of
the hypotenuse
b, The length of the longest leg is the length of the shortest
leg times 3.-
4. Consider the triangle below.
a. If c = 20, then b = 1 '2
b. b
c. a =
RIGHT TRIANGLES
Suppose we have an isosceles right triangle as pictured below with the lengths
of the two equal legs x.
The length of the hypotenuse is found in the following way.
x2 x2 = c2
2x2 = c 2
x = c
This means that the length of the hypotenuse of a 45°-45°-right triangle can be
found by simply taking the length of a leg and multiplying by the 45--.
5. Consider the triangles below:
a. Find c.
C
b. Find x
7
25
POLYGONS AND THEIR AREAS
Polygons are plane figures formed by line segments which meet at their end-
points. They are named according to the number of sides they have. Some of
the common ones are shown below.
Triangle: 3 sides
Quadrilateral: 4 sides
Pentagon: 5 sides
N-gon
11. ,t(10,
Octagon: 8 sides
Decagon: 10 &des
POLYGONS AND THEIR AREAS
Let's try to develop an easy rule to find the total number of degrees contained
in the sum of the measures of all the angles in any polygon. A basic premise we
work with is that every triangle has a sum of 180°. With this knowledge,'divide
each of the polygons pictured on page one into as many distinct triangles as possible.
With each, work from one vertex only. Fill in the table below with information gained
from the above activity.
Type ofPolygon
Sides A'sFormed
Degrees inEach 6
Total Degrees inthe Polygon
Triangle
Quadrilateral
Pentagon
Hexagon 6 4 180 4 x 180° or 720°
Octagon
Decagon
N-gon N 180
To be able to fill in the last row for N-gon we must see the relationship between
the number of sides and the number of triangles formed in each polygon. In each, the
number of sides minus the number of triangles formed is . There-
fore, the number of triangles formed by the N-gon would .be N - 2. The total number
of degrees in the N-gon would be: number of Q's formed times number of degrees in
each P or (N -2) X (180).
POLYGONS AND THEIR AREAS
This formula can be used to find the number of degrees in any polygon by substituting
the number of sides for N.
I. Find the number of degrees in a seven-sided polygon (heptagon). degrees.
A regular polygon is one that has equal length sides (equilateral) and equal measure
angles (equiangular). Draw a regular quadrilateral. Give the common name for this
figure.
3
POLYGONS AND THEIR AREAS
2. Knowing how to find tlic: number of degrees in any polygon and knowing the
definition of regular polygons, try to find solutions to the following
questions.
a. Find the number of degrees for each interior angle of a regular pentagon.
L= 0
b. Find the number of degrees for each interior angle of a regular octagon
0
The formula to find the number of degrees in an interior angle of any pol-
ygon is in-2)180 = sum of all anglesn number of angles
4
POLYGONS AND THEIR AREAS
Relating Regular Polygons to Circles
To construct any regular polygon we may make use of the circle. For example,
construct a regular pentagon utilizing a circle. Use a compass and protractor for
the constructions.
Step 1. Construct the circle (See figure below.)
Step 2. Divide the circle into five (pentagon) equal arcs. Each would have a
central angle of 5 360° or 72°. (A central angle has its vertex at
the center of the circle.)
Step 3. Connect the five points.
3. What measure of central angle would be utilized to construct an octagon?
Construct an octagon using the circle method.
5
POLYGONS AND THEIR AREAS
Finding Areas of Polygons
Because all polygons can be divided into a certain number of distinct triangles,
we can find the area of any polygon by finding the sum of the area of these
triangles.
Formula for the area of a triangle
A = 1 bh2
b = base
h = height to base b
Formula for area of a triangle knowing the lengths of the three sides.
A = 11S(S-a)(S-b)(S-c)
S = 1 (a+b+c)2
4. Area =
6
5. Area =
POLYGONS AND THEIR AREAS
To find the area of the irregular pentagon
to the right, we can divide it into triangles
whose areas we can find.
6. Find the area of the irregular figure to the right.
Area =
Common Polygons That Have Specific Formulas for Their Areas
Parallelogram (two pair of parallel sides)
A =bxh
b = AB
h = DE
Trapezoid (only one pair of parallel sides)
A = 1 (b1 x b2)h2
b1
= length of top base
b2 = length of bottom base
h = 1 height = AE
7. Find the area of the trapezoid given above.
Area =
7
1)
bx= 10
POLOYGONS AND. THEIR AREAS
The area of any regular polygon can be
found by finding the area of one of the
triangles, then multiplying by the number
of sides. In this case the height of
6ABO is OX.
Area of a regular polygon = N a bh)2
Where
N = number of triangles
b = base of one A = AB
h = height of one4. = OX
8. Find the area of the regular octagon where OX = 10 and AB = 8.
Area =
8
Independent School District 281
PARALLEL LINES
Definition: Two lines are parallel if they lie on the same plane and dc not
intersect.
(r-
..11 is parallel to 12
Many problems exist involving parallel lines which are intersected by a third
line. We call this line that intersects the two parallel lines and is in the
same plane as the two parallel lines a "transversal."
ti
In the above figure, line t is a transversal cuttingliand .172 at the two iaints
A and B. The transversal must cut the two lines in two points (or three lines in
three points, etc.).
Keep in mind that a transversal can intersect any, two or more lines; however,
in this discussion we are only concerned with what happens if we intersect
parallel lines with a transversal.
PARALLEL LINES
1. Consider the following parallel lines cut by a transversal
Use your protractor to find:
a. m L 1 = e. m L 5 =
b. m 4 2 = f. m L 6 =
c. m L 3 =g' m 47 =
d. m 4: 4 = h. m 4 8 =
Aq2
transversal
The angles named in the figure above are given "special names in pairs."
2. The pair 4: 3 and Z. 5 are called "alternate interior" angles. There is one
more pair of alternate interior angles. What is it? a.
and b. How do the measures of a pair of alternate interior
angles compare? c.
3. The pair 1 and Z. 5 are called "corresponding" angles. Another pair of
corresponding angles is 4 and 48. There are two rore pairs of correspond-
ing angles. What are they? a. and b. ; c.
and d. . How do the measures of a pair of corresponding angli:
compare? e.
2
PARALLEL LINES
4. Angles 4 and 5 are supplementary (i.e., m L. 4 + m AC 5 = 180). Name
another similar pair of supplementary angles. a.
and b.
There are other important properties of the angles, however the three types mentioned
in problems two, three, and four above are the situations we will concern ourselves
with mostly in the units to follow. Remember, the following facts. Anytime you cut
parallel lines with a transversal,
a. The pairs of alternate interior angles have equal measures.
b. The pairs of corresponding angles have equal measures.
c. The interior angles on the same side of the transversal are supplementary.
5. Check your results in problem one with the summary just stated. Do you find this
true in problem one? If not, then you better
re-measure your angles.
There are some interesting and useful properties concerning parallel lines cut
by transversals.
Property 1. If a line is parallel to one side of a triangle and contains the midpoint
of a second side, then the line contains the midpoint of the third side.
3
If." contains midpoint of AB,
jis parallel BC.
Conclusion: se contains midpoint
of AC
PARALLEL LINES
Property 2. If parallel lines cut off\ congruent segments on one transversal,
they will cut off congruent segments on any other transversal in-
tersecting them.
If /a # .f3 and a = b = c
Conclusion: r= s= p and x= y= z
Property 3. If parallel lines cut two or more transversals, then the lengths
of the segments intercepted pn one transversal are proportional
to the lengths of the segments intercepted on the others(s).
4
PARALLEL LINES
If 4 44.11 13 and 4/andttare transv,:!rsals.a c a , b
Conclusion: b d or c d
Property 4. If a line intersects two sides of a triangle and is parallel to
the third side, then a triangle similar to the original triangle
is formed and the ratios of the corresponding sides of the two
similar triangles are equal.
6.
f3
5
If DE //
Then L ADE /1 ABC and
a b e
AB AC c
Given 1; -121/13
Find x =
PARALLEL LINES
If 11 and AG intersect at A,
and AB = BC = CD = DE = EF = FG,
describe a method for partitioning
AP into six congruent segments.
Could you now partition AB into six congruent segments? Triple
the lengths of the picture shown with the same angle and show all
lines (lightly) that divide AB into six equal segments.
6
STANDARD CONSTRUCTIONS
In this unit ycu will have the basic constructions at your fingertips. No proofs
of any of these are given. The purpose is rather that the student be able to
have a short list of the constructions more commonly used and perhaps a few
others.
1. Construct a triangle, given the three sides a, b, and c.
Procedure
a. Make AB = c.
b. Using A as a center and b as a radius,
draw an arc.
c. Using B as a center with radius a, draw
another arc which will intersect the
previous arc at a point called C.
d. Q ABC is the required triangle.
2. Given an angle, construct its bisector.
Procedure
a. Using B as a center and any radius,
draw an arc which intersects angle
sides at D and E.
b. Using D and E as centers and
equal radii, draw arcs which then
intersect at F.
c. BF is the required bisector.
b
C
STANDARD CONSTRUCTIONS
,3. Duplicate an angle (G. CDE) with a Oven vertex (A) and side (AB).
Procedure
a. Using D and A as centers, draw equal
arcs, intersecting L's at F, G, and K
respectively.
b. Using FG as a radius and K as a center,
draw an arc which will intersect arc
HK at M.
c. Draw AM.
d. MAB is the required angle. .1
H
1
A K 8
4. Draw a triangle, given two sides and the included angle.
Procedure
a. Construct AB equal to c.
b. At A, construct de BAT' = 4:1.
1111111
c. On AF, construct AC = b.
d. 21 ABC is the requiredia.
6
5. Construct the perpendicular bisector of a segment.
Procedure
a. Using A and B as centers and using
equal radii, draw arcs intersecting
at C and D.
ems.b. CD is the required 1 bisector of
AB.
2
A
CF
\
A
C
1
STANDARD CONSTRUCTIONS
6. Construct a perpendicular to a line from an external point P.
Procedure
a. With P as center, draw an arc which
intersects the line at points A and B.
. 15. Using A and B as centers and equal
radii, draw arcs which will intersect
at C.
c. PC is the required perpendicular.
7. Construct a perpendicular to a line at a particular
Procedure
point (P) on that line.
a,
b.
Using P as center, draw a circle in-
tersecting the line at A and B.
With A and B as centers and equal
radii, draw arcs which intersect at A p 'B
C.
c. PC is the required perpendicular.
8. Through a given point (P), construct a line parallel to a given line.
Procedure
a. Through P, draw any line which
intersects the given line L.
b. Using P as vertex and PB as a
side, construct 4,3T =Atx.
c. L' is the required line.
3
STANDARD CONSTRUCTIONS
9. Divide a line segment into any number of equal parts.
Procedure (given for three)t": C
a. From A, draw a line other than It ,E,)<_
s N.
.
b. Construct three equal segments: AD, DE, K)..
sD,K ...
.%. ...- -, ,,
. `Band EF, on that line. A H G
cC. Draw FB.
d.
e.
Through D and E, construct lines parallel to
FB, intersecting at H and G respectively.
AH = HG = GB
10. Bisect an arc AB of a circle.
Procedure
a.
b.
c.
Draw chord AB.
Construct CD, the perpendicular bis-
/1ector of XE, which intersects AB at E.
Then XI =
11. Find the center of a circle, given only an arc.
Procedure
a. Draw two distinct chords whose endpoints are on the arc.
b. Construct thel - bisectors of both chords.
c. Their intersection is the required center.
4
'Ds
STANDARD CONSTRUCTIONS
12. Construct a tangent to a circle at a given point (A).
Procedure
a." Draw a line from 0 through A.
b..11w
Construct a perpendicular to OA,
through A.
c. This perpendicular is the required
tangent.
13. Construct tangents to a given circle from a given external point.
Procedure
a. Draw 31
b. Bisect 17, calling that point B.
c. With B as center and OB as radius, C/
oLAdraw a circle which intersects the
\given circle at C and D.------
\ D(Nd. AC and rare the required tangents.
t-li
14. Inscribe a square in a given circle.
Procedure
a. Draw any diameter AOB.
b. Construct a JL- bisector of AB,
intersect at C and D.
c. ADBC is the required square.
STANDARD CONSTRUCTIONS
15. Inscribe a regular hexagon in a given circle.
Procedure
a. Draw any radius OA.
b. Using A as center and radius OA,
draw an arc which intersects the
circle at B.
c. In like manner, using B as center,.
find C; then D, E, and F.
d. ABCDEF is the required hexagon.
16. Circumscribe a circle about a given triangle.
Procedure
a, Constructi-bisectors of any two sides
of the 41.
b. Using the intersection point as center 0,
and OA as a radius, construct the required
circle.
17. Inscribe a circle in a given triangle.
Procedure
a. Construct bisectors of any two angles\
of the triangle and call the intersec-
tion point O.
b. From 0, construct a perpendicular to side
AB and call the intersection point F.
c. Using 0 as center and OF as the radius,
draw the required circle.
6
STANDARD CONSTRUCTIONS
18. Divide a line segment into n parts which are proportional to n given
line segments.
Procedure (shown for three parts)
*-11,
a. Draw AC , distinct from AB.
b. On AC , construct AD = m, DE = P,
and EF =
c. Draw FB.
d. Through E and D, construct lines parallel
to FB, intersecting AB at G and H
respectively.
e. Thus S = =m p
Ax H u G z B
;\:
7%-", \P
'it:. /N-E
r ,
19. Construct a polygon similar to a given one when two corresponding sides
are given.
Procedure (ABCDE is given: A'B' is given)
a. From A, draw all diagonals.
b. At A', construct x' =4..x and extend
side.
c. At B', construct B' = L.B. The
intersection is C'.
d. In like manner, constructi..mt4m,
= hy, and call the intersection of
the sides D'.
e, Also in like manner, constructAn'
1.pl =1,p, to find intersection E'.
f. A'B C'D'E' is the required similar
polygon.
7
GF
1I
A'
STANDARD CONSTRUCTIONS
With the basic constructions given here, you should be 'able to figure out how
to construct all of what you would need in the various on-the-job situations.
you'll encounter. On separate paper, try the following constructions to see
if you've mastered the techniques shown above.
1. Divide the base of a triangle into segments proportional to the
other two sides.
2. Through the vertices of a triangle, construct parallels to the
opposite sides.
3. Construct a 30° -60° right triangle and inscribe a circle in it.
Circumscribe a circle about the triangle.
4. Given a circle and the midpoint of a chord, construct the chord.
8
VOLUME
The cubic yard, which is an application of the volume principle, is used consistently
in construction work. The relationships between cubic inches, cubic feet, and cubic
yards are shown below.
ONECUBICINCH
ONEFOOTJOR 12 IN
fig. B
"'ONE FOOTOR 12 IN.
ONE CUBIC FOOT =1728 CUBIC INCHES
O
z0
LONE YARD. OR 3 FT,
ONE CUBICYARD = 27
CUBIC FEET
-"ONE YARDOR 3 FT.
II. Volume of a triangular prism
V =Bxh
The base is a right triangle with sides of 3",4",5". Find the area of this
triangle. (See unit "Polygons.")
B = /s(s- a)(s- b)(s -c) wlaere s = 1 (a+b+c)\ [6, b, c are the lengths of the sides.)2
s = i (3+4+5) = 6; B = V6(6-3)(6-4)(6-5) = V/6(3)(2)(1) = 1/777= 6 sq. in.
2
h = 4 inches
V = 6 x 4 = 24 cubic inches
fig. C
4 id
VOLUME
III. Volume of a pentagonal prism
V =Bxh = area of the base
Step 1. Find area of the base.
Step 2. Find height.
Solution:
Step I. Notice the base can be divided into a rectangle and a triangle...
Find the area of the rectangle, then the area of the triangle.
Area of rectangle = 8 x 10 = 80 square inches.
Area of a triangle = 6(s-a)(s-b)(s-c).
Where s = 1 (a+b+c)4
S = 1 (8+8+10) = 132
Therefore A = Y13(13-8)(13-8)(13-10) = /13(5) (5) (3) = 4777 = 38.4
Total area of the base = 80 + 38.4 = 118.4 square inches
Step 2. Height = 4 inches
Volume = B x.h = 118.4 x 4 = 473.6 cubic inches.
fig. D
10"
4
h711
VOLUME
IV. Volume of a trapezoidal prism (a prism whose top and bottom bases are
trapezoids).
Notice the trapezoidal base is not
in the bottom position, but on the
side; even so, it is, by definition
the base of the prism because the
opposite side is the same size and
shape.
V = B x h
B = 1 (b1 + b2) h2
=1
2(14' + 6')16'
= 10.16
= 160 square feet of a trapezoid.)
fig. E
(See unit, "Polygons," fc,r explanat1,1 of area
V = 160.18' = 2880 cubic feet = a. cubic yards. (Find how \many
cubic yards)
Find the volume of dirt to
be removed in figure F.
Give your answer in cubic
feet and cubic yards.
AF = 60' i4
MB = 40'
fig. F
/0
5
40'
cubic feet = b.
VOLUME
cubic yards = c.
V. Volume of a circulal. cylinder
Note: Circular cylinders are not prisms as they do not have polygons for
bases.
Formula: V = Bxh
In this case B, the area of the base,
is the area of a circle.
Example:
Find the volume of this cylinder.
V =B zh
= 432x 922 9 9= x x7 I I
_ 1782
\7
254.56 cubic units
Volume of fig. I =
fig. G
fig. H
9
3
fig. I.
IS
CIRCLE RELATIONSHIPS
For the purposes of this unit, a circle shall be indicated as C(P,r), which will
signify a set of points equidistant from a particular point P. P is known as
the center ar7.. r will be called the radius. Any segment originating at P and
having length r may be used as a radius.
Definition: Two or more circles are
concentric if and only if they have
the same center. In the diagram,
circles C(P,R) and C(P,S) are concentric
circles.
Definition: A chord of a circle is a segment
(i.e., AB) whose endpoints are points of
the circle. A diameter is the measure of
the chord which contains the center of the
circle; when a plumber talks of a 3inch
pipe, he's referring to diameter.
Definition: Congruent circles are circles
with congruent radii.
Fact 1: Radii of congruent circles are congruent.
Fcct 2: The diameter of a circle is twice the radius of that circle,
Fact 3: Diameters of congruent circles are congruent.
Definition: The interior of a circle is the set of points in the plane cf the
circle whose distances from the center are less than the radius.
CIRCLE RELATIONSHIPS
Definition: The exterior of a circle is the set of point's in the plane of the
circle whose distance from the center are greater than the radius.
Definition: A line coplanar with a circle is tangent to .//'
the circle if and only if its intersection with the
circle contains exactly one point. For example, RA
i. tangent to C(P,R) at R, the point of tangency.
1. In the following three circles, give examples where the intersection of a line
and a circle contains 0, 1, and 2 points. Is it possible for the line and circle
to intersect at more than.two points?
a. b. c.
Definition: Two circles are tangent if and only if they are coplanar and are tangent
to the same line at the point. Circles may be tangent either internally or
externally.
CIRCLE RELATIONSHIPS
Fact 4: A line coplanar with a circle is tangent to the circle if and
only if it is perpendicular to the radius whose outer end is the
point of contact.
Fact 5: If two circles are tangent, their centers and the point of contact
are collinear.
2. Given three circles each tangent to the other two, there are four poss-
ible arrangements, one of which is shown. Sketch the other three.
a. b.
,Definition: A line tangent to each of two circles is called common tangent. If the
line intersects the segment connecting the two centers, it is called an internal
tangent; if not, then it is an external tangent.
17 is an internal common tangent
CDr is an'eXternal common tangent
3
CIRCLE RELATIONSHIPS
3. Sketch examples showing two circles having 0, 1, 2, 3, and 4 common tangents.
a.11 o 11 b.
d. "3" e. 11411
c. 11211
Many facts involving chords and their relations in circles are useful.
These will be used later.
Fact 6: A line containing the center of the circle that bisects a chord
other than a diameter, is perpendicular to that chord.
Fact 7: A line containing the center of a circle perpendicular to a chord
bisects that chord.4
Fact 8: In the plane of.a circle, the perpendicular bisector of a chord
passes hroughtthe center of the circle.
Fact 9: Chords of congruent circles are congruent if and only if they are
equidistant from the centers of their respective-circles.
4
CIRCLE RELATIONSWIPS
Definition: A central angle is an angle in the plane
of the circle which has the center of the circle as its
vertex. If APB is a central angle.
Definition: The measure of any arc shall be the same as the measure of the
central angle intercepti9g that arc.
Fact 10: Arcs of congruent circles are congruent if and only if they, have
Bthe sal,.,:: 4,aasure.
Fact 11: m AXB = m AX +.cirCB
Definition: An angle is inscribed in an arc if, and
only if, each side contains\one end point of the
arc and the vertex of the angle is a point of the
arc other than an end point. The arc of the cir-
cle whose points lie either in the interior of or
on the sides of the inscribed angle is called the
inscribed arc.
Fact 12: The measure of an inscribed angle is half the measure of its
intercepted arc.
Fact 13: The measure of an angle having a chord as one side
and a tangent as the other is half the measure of
its intercepted arc.
5
CIRCLE RELATIONSHIPS.
Definition: A secant of a circle is a line which
intersects the circle in two distinct points.
Fact 14: The measure of an angle with its vertex in the interior of a cir-
cle and whose sides are contained in secants to the circle is half
the sum of the measures of its inte--epted arc and the arc inter -
depted by its vertical angle.
Fact 15: If .9.n angle intercepts two arcs of a
circle, its measure is half the difference
of the measures of the 'arcs.
Fact 16: Angles inscribed in the\same or congruent
arcs of a circle are congruent.
Definition: Given a tangent from external point A to a circle, the tangent segment
is the segment with endpoints A and the point of tangency and all points of the
tangent between these two.
Fact 17: The two tangent segments from a point
in the exterior of a circle are con-
gruent. AB AC
Try the following:
CD\ 4. If m BC= 120, m = 90, and m CD = 60, then:
a. m 1 =
b. m Z. 2 =
c.
d.
_ 3m ."1
Ts L4
=
=
6
CIRCLE RELATIONSHIPS
5.. If rr. / 1 = 40 and m L 4 = 25, then
a. m = b. m CD =
6. If m AC = 110, and n R3 = 36, then
M Z. ARC =
7. If m CA = 110, then
m L ARC =
Fact 18: The length .4.1 of an arc of radius r and measure m,
Definition: A sector of a circle is a region bounded
by two radii and an arc as shown in the diagram.
180 (Irr)
Fact 19: The area of a sector of a circle is half the product of its radius
and the length of its arc.
1area of sector = x r2
1-Txrx180 (11.)
= m- (17'r2)360
Definition: A segment of a circle is a region bounded by a chord.
To find the area of a segment, first find the
area of the sector and then subtract the area
of the triangle contained in the sector.
Finding the area of the triangle may be a bit of a job, especially if the
mill11111101111pr
central angle associated with the segment does not have a measure of 600,'900, or
1200. In this case, you'll be able to make use of the 30-60-90 triangle, the
CIRCLE RELATIONSHIPS
isosceles right triangle, or the equilateral triangle relationships. In other cases,
you'll need to use numerical trigonometry. An example of how such a problem would
be set up is given in the accompanying diagram:
AB is the radius. Draw BC AD.
Then 4: ABC = 1 the central angle.2
sin B = h; cos Bc
a
c
Then, use the formula for the area of a triangle.
11111111111101000
8. Find the lengths of these arcs.
a r = 10, m 40,i= c. r = 8, m = 90,..Jt=
r = 25, m = 424= d. r = 500, m = 1504=
9. Find the area of the' sector of a circle with radius 8 whose arc has the
following measure.
a. 60 C. 90
b. 150 d. 160
10. Find the area of the shaded region where given:
a. m Z. P = 9U, r = 12,
b. m L P = 60, r = 4,
c. m 4_ P = 100, r = 6,
d. m L P = 120, r = 2,
CIRCLE RELATIONSHIPS
For a little variation, try the fcllowing:
11. Triangle ABC is equilateral and the radius of eacE arc is 12, find the area of
the shaded region.
C
II. CA
RE
ER
UN
ITS -- T
AB
LE
OF C
ON
TE
NT
S
1.
Printing and Graphic Arts
19.
Plumbing and Pipe Fitting
2.
Heavy Equipment Operator
20.
Surveyor
3.
Fashion and Apparel Design
21.
Outdoor Advertising
4.
Navigation
22.
Space
5.
Painting and Paperhanging
6.
Landscape Technology
7.
Carpenter
8.
Architecture and Drafting
9.
Optical Technician
10.
Sheet Metal
11.
Engineering
12.
Machinist
13.
Cement Worker
14.
Forestry
15.
Electrician
16.
General Contractor
17.
Home Planning
18.
Cabinetmaking
PRINTING AND GRAPHIC ARTS
Al! of us have looked through the telephone book; few of us, however, have
stopped to think of all the work, clone by many different kinds of people, that
has Fone into the books. Even after the copy (material that goes on the pages)"
is ready, it takes much effort by many people to get the book prepared. Most
of this work is done by people from the printing industry.
There may be as many as 25 different specialized jobs in this general industry,
each with its owl. requirements, pay, opportunities, etc. Generally, however,
the workers can be divided into three areas: printing craftsmen, preparation
and management workers, and the allied skilled workers. The machine operators,
pressmen, cameramen, etc., make up the craftsmen category. People ranging from
reporters to accountants and executives make up the next category, and all of
the other workers who help in the actual production of the material would fit
into the area of allied skilled workers.
Most workers become skilled by going through an apprenticeship program. It may
be very formal as is the case with most typesetters or it may been informal
program of on-the-job training as in the case for an artist or reporter. Another
means for obtaining training would be to attend a vocational trade school. Many
courses and programs are available and in some (the area vocational schools)
there is no tuition if you are under 21 and a high school graduate.
The printing industry (graphic arts) is involved in the preparation of any kind
of written material. The telephone book mentioned earlier, the textbooks we
use, the newspaper, etc., are all prepared by some phase of this industry. Many
of the workers become involved in math very quickly. For instance, in working
with the printing of books, the type page (that is, the part of the page the
printing covers) which will appear to be the most pleasing to the eye should be
1
PRINTING AND GRAPHIC ARTS
equal to one half the area of the paper page. Another way of stating this rule
is that the printed part of the page should equal the area of the white marginal
space surrounding it and the sides of the type page are to be proportional to
the :sides of the paper page.
Most book printing is in the form of a rectangle. Those rectangles most commonly
used are:
Those rectangles most commonly used are:
Rectangles Root-2 De Vinne Golden Root-3 Root-4
Factor 1.40 1.50 1.60 1.70 2.00Reciprocal .71 .75 .78 .82 .87Type area ( %) 50 55 62 67 75
Margin area ( %) 50 45 38 33 25
Root-1
Basic unitof design
Root-2
I 1 1 I
De Vinne Golden
I 1 I
Square Hypotenuse Regular1:1 5:7 4:6
Golden3:5
Root-3 Root-4
Printers Long2:3 1:2
According to the table, which rectangle (notice the root-2 rectangle is also
called a hypotenuse rectangle, etc.) most closely meets the rule on what is the
most pleasing ratio of type area to margin area? 1.
2
PRINTING AND GRAPdIC ARTS
If all you are interested in is getting the most type area on the paper, which
oblong would you choose? 2. . Measure your
geometry book (length and width) anck determine which rectangle the publisher used
3. . If a b?ok measures. 10" by 6", which. rectangle is
used?- 4. . One measuring 15" by 21" follows the
5. oblong.
The following is an example of a scale layout..
I-
This would be a layout for a golden rectangle.
Using the same scale, make a layout for each of the following paper page dimensions
and tell what type of printer's rectangle they fora?.
6. A paper page 9" by 6" 7. A paper page 21 by al.2 2
PRINTING AND GRAPHIC ARTS
The rectangle used most often is the root-2 or hypotenuse rectangle. We can
coastruct a root -2 rectangle by starting with a square. Follow the
directions to construct a root -2:
A. Construct a square.0
B. Draw the diagonal.
LOC. Raise the diagonal to a vertical
position -- this gives you the
height of the root-2 rectangle.
Finish the rectangle using the height
dimension.
I.0
8. Start with a three-inch square and construct a root-2 rectangle. Begin the square
low enough to allow for the diagonal.'
Ir
PRTNTING AND GRAPHIC ARTS
To help determine where the typed material should be on the page, printers have
three centers to judge with.
50%
The moth center = 1 of page length9
The opticalcenter = .45 of page length measured fromtop of page
The line of golden proportion = 2 of page length8
. measured from top of page
50%
MATHCENTER
45% 3
550/0
OPTICALCENTER
5
LINE OF GOLDENPROPORTION
Notice Oe optical center (the areas of the page our eyes center on) is dif-
ferent from the math center.
9. Find the optical center of the-root-2 rectangle in problem eight. (Use the
diagram.) Answer:
The composer may take the optical' center into consideration and leave more
margin on the bottom and extend the type to the top. Is this true with any of
your school hooks?
Example
SINGLE PAGE2:3 RATIO
When a composer wishes to enlarge or reduce a drawing, photograph, etc., he
will often use the diagonal line method. The principle that rectangular areas
enlarge and reduce proportionally along their diagonals is used.
5
PRINTING AND GRAPHIC ARTS
4
// ;
DIAGONAL LINE METHOD
One dimension of the enlargement of 'reduction must be known. For example, a
3 x 5" (width x depth) picture is to be enlarged to an 8"-wide copy.
5"
/
Scale I", = 1"4
To find the remaining dimension of the enlargement (X) we would use the proportion
formula on page 7.
6
PRINTING AND GRAPHIC ARTS
FORMULA: width of original = desired widthlength of original desired length
thus 3" =511
3x = 40 (cross multiplication)
x = 40 or 131"3 3
10. Find the remaining dimension of an enlargement of a 4" x 7" picture enlarged
to a width of 11". Make a diagram similar to the example above with a scale of
i" = lu.4
The previous material gives a sampling of the type of geometry used in graphic
arts careers. There are other math applicationsused in this trade that are not
covered in this unit. Please feel free to ask your teacher or counselor for
more information.
7
HEAVY EQUIPMENT OPERATORS
Heavy equipment operators work with many different kinds of construction equip-
ment such as cranes, grders, bulldozers, etc. Over 310,000 of these operators
are employed throughout the country and earn between $5.25 and $7.04 an hour.
Most of these operators have completed an apprenticeship program consisting of
up to three years of on-the-job training with some (144 hours) related class-
room instruction. This related instruction is usuallly in areas such as equip-
ment maintenance, electricity, welding, reading plans, etc.
Other operators attend school full time at a local trade school or receive the
training while. in the armed forces. In Minnesota, the Staples Area Vocational
School provides a program in heavy equipment operation and maintenance that
takes twenty-one months.
The government predicts that the number of equipment operators needed in the
near fut".re will increase rapidly. As construction activity increases and as
highway construction continues, job opportunities will also increase. Larger
and more efficient equipment will tend, however, to offset some of these increases.
Operators need good hand-eye coordination and judgment plus mechanical aptitude
and an interest and desire to work physically in the out-doors.
The heavy equipment operator must also have a complete understanding of blue-
prints as well as the ability to visualize the structure to be built.
He must also know the nature of different types of soils.
HEAVY EQUIPMENT OPERATORS
An example of a diagram of a typical job is the basement plan for a bank below.
Fig. 1
O
DOILLtz, c1.00%4 STORA CACVNULT
V49 wilta014
4, 0
STO,q
/WAINWWWW
2UNIT IOTENILK
1. SNT.
C.A.c
4,-0"
]PACE. AQGA"N
13'- 71 &-e
F14 Cl 61..r..V. '4
0.100.1rea,5700.hGli
ROOM
AP.rtll
BASEMENT PLANSCALE 1/8" 1, -0"
The walls of the basement are to be block, so footings for the blocks
must be parallel. The equipment operator must dig a trench for these foot-
ings. They are 12" wide and 10" deep and will go around the entire building;
the walls of the basement will go 8' into the ground. In order to "price"
this job, the total volume of dirt to be removed must be found. The estimator,
2
HEAVY EQUIPMENT OPERATORS
working with the equipment operator, had to find the following information.
1. Total area of basement =
2. Total volume of basement =
3. Length of footings = (Total)
4. Volume of footings = length x width x height =
5. Total volume of basement and footings (Give answer in cubic yards.) =
HINT: 27 cubic feet = 1 cubic yard
With this information, the correct number of trucks can be arranged to carry
away the dirt. An accurate estimate of the cost can also be completed.
Many people who are making improvements around their homes act as their own
general contractor. These people oftentimes rely on the knowledge of drivers
for trucking ,companies to help them make decisions. For example, Tom Jones
had a crushed rock driveway and wanted to concrete it. (See fig. 2.)
3
Fig. 2
HEAVY EQUIPMENT OPERATORS
54"0"
Tom called the G & P dirt removal company to remove the crushed rock and go
as deep as necessary for the concrete to be poured. They informed Tom that
they charge $45 per hour for the use of their machines and labor, with a
minimum charge of $50 for any one job. G & P also said that they can remove
roughly 10 cubic yards of dirt per hour. Tom's next job as his own general
contractor was to find the number of cubic feet in his driveway. He wanted
to go 6" deep to allow for 2" of sand and 4" of concrete.
He drew a diagram of his Y-turn driveway
similar to figure 2, added measurements,
and divided it into two areas -- which
resulted in figure 3. The area of B (see
figure 3) would be that of a rectangle, so
Tom had no trouble finding its volume. The
area of A, however, was oddshaped. Tom de-
cided to block off a rectangular area (IJKL),
find the area of it, then subtract the area
of the two partial circles (notice each is
one-fourth circle) called sectors of a circle.
4
HEAVY EQUIPMENT OPERATORS
See if you can find the following needed information.
6. Area of B
7. Area of IJKL
8. Area of one of the sectors
9. Area of A (answer to 7 minus the answer to 8 doubled)
10. Total area of driveway
11. Total volume of driveway (6" deep) in cubic feet
12. Total volume in cubic yards
With this information, Tom knew how much it would cost to remove the crushed
rocks and dirt. (Remember the charge is $45 per hour and that they can remove
10 cubic yards per hour.) The cost would be 13.
5
HEAVY EQUIPMENT OPERATORS
The next day G & P sent out a couple of men and a backhoe-loader combination.
The backhoe would dig out the old rock and dirt and.the loader (on the other
end of the backhoe)'would scoop up the loosened debris and put it in the truck.
The operator said the backhoe itself could hold up to one-half cubic yard of
material, but because of the hardness of the soil, it usually held one-fourth
cubic yard. The loader', scoop had a capacity of one cubic yard. The operator
suggested to Tom that he order the sand as soon as possible. Already knowing
the area of the driveway and that 2" of sand is needed, find the number of
cubic yards of sand needed. (Hint: determine the volume of sand needed by
multiplying area x depth in common units. Then change to cubic yards.)
14. Cubic yards of sand needed =
This same operator has handled some of-the big cats, graders, and earth movers
seen so often on highway projects. He said that heavy equipment operators
must work their way up from small to large projects in steps. Experience is
the key to success!
FASHION AND APPAREL DESIGN
Fashion and apparel design is an occupation that lends itself readily to a
person with an artistic ability combined with creativity and an adventuresome
spirit. Designers set trends in clothing styles and sell those "looks of the
future" to the public. It is a demanding career, for if one works merely upon
what is currently in vogue, then consumers will soon realize this and begin
buying from one's competitors.
A fashion designer creates original designs for new types and styles of apparel.
The creation of the new styles occurs in steps; first, a sketch is made, then
approval is given by management, and finally an experimental garment is made.
The designer is almost always directly involved in all of these steps. There-
fore, an artistic ability, and ability to "sell" your new idea, and an ability
to actually make the product are all important for the designer to have.
As is true for most jobs involving creativity, the designer will probably
have much freedom. She/he may travel in order to pick up new ideas and much
time and effort is consumed before definite styles are developed. The pressure
to produce is present however, even with this freedom, because it is easy to
judge how well a designer is doing by the number of successful new styles
developed over a period of time. Style must be developed several seasons and
sometimes a full year ahead of the production of the garment, so ideas must
be several steps ahead of current fashion.
Much like other creative fields, the prospective designer enjoys the possibility
of great reward -- both in fame and in fortune. The number of designers who
"make it" to this point is very small compared to the many who aspire. Many
would-be designers end up in related fields because the competition is so great
in the area of design. These related fields include wholesale buying for stores,
FASHION AND APPAREL DESIGN
copying of successful designs, sales or advertising, or maybe even working in
the actual production of a garment.
Most of the basic ideas covered in this unit would be covered in much more depth
in any sewing course. When this is studied in a home economics course, one
doesn't realize the large amount of mathematics which is involved.
For example, two of the basic pieces of equipment which will be used by anyone
seriously interested in tailoring are a pressing cushion and a pressing edge.
The pressing cushion is needed to help press darts in fabric pieces of the
blouse and skirt to create curved lines which will drape more attractively over
the body, whereas the pressing edge is used to press open new seams along their
line of stitching without disturbing the remainder of the fabric. Using the
scale given below, what is the length I. and width 2.
of the pressing cushion?
The pressing cushion can be made and
filled to a solid pack with wool
batting or sifted sawdust.
2
Length
(I square =.1 inch)
FASHION AND APPAREL DESIGN
3. Using the scale given for the pressing board, draw the board to the proper size.
(I square = I inch)
I_
_.4,
i i I 7 I--
PRESSING EDGE. Board 5" x 12"
1 I
BASE I" Board 5" x 12 '
Glue and fasterwith two 1;2 x'10 flat -hecc
wood screws from bottom
The pressing board can be handmade by using a one-inch board, 5" by 12", and
cutting a long tapered point on one end as shown above. The narrow edge is
fastened upright (with two screws) to a second board, 12" long, to form a
stand. The end with the tapered point is used for pressing.
As you can see, most diagrams involve proportions, such as one square = one inch
as used here. Later, you'll have the chance to do more "blow-ups" of simple
patterns.
An area of geometry not studied too much at first is called translations. It
basically means having a pattern of some sort duplicated again and again along
a straight line, such as .44/p1 Jove , etc. A type of transormation
results in a series of parallel lines , which may be thought of as
lines which are "translated" along a vertical line.
3
FASHION AND APPAREL DESIGN
You're probably asking just what all this has to do with tailoring. Frankly,
a lot, for this type of geometry is applied in the most fundamental ingredient
of good tailoring, the stitching. All stitches have a purpose, and, if you
examine good stitching closely, you'll find that it is all composed of repetitive
patterns, or translations. Notice the translations in the following common
stitches.
A.
Blind Hemming
D.
Backstitch
B. E.
Catch stitch
C.
Blanket Stitch
4
Floating or Slipstitch
F.
Padding stitch
FASHION AND APPAREL DESIGN
G.
eoe eciA,
Stab Stitch
H.
Tailor Basting
I.
Catchstitch
Can your eyes always believe what they see? Through the clever use of optical
illusion, it is possible to deceive the eye. There are many examples of this
fool-the-eye magic. Here are some of the more familiar.
Which line is
longer? Of course,the lines are thes:line length, hutthe optical illusionmakes you thinkotherwise. Deflect-ing the gaze down-ward in -A seems toshorten Coe line.'Extending the lineas in B makes theline seem longer.
/\
\./A B
Is A longer than B? No, lint the longerline under B makes B look shorter.
B
Arc the verticallines parallel? Yes.The zig-zag of theopposing lines onlymakes them appearto waver.
5
Do the verticallines bulge in thecenter? Although itworld appear so,this is merely anoptical illnsioi.i. Thelines are parallel,but the many linesthat converge uponone point call atten-tion to that area,making it appearwider.
Which line is longer, A or B. Both arethe same length, but A seems to he longerbecause the line above it is the same length.A longer line or added width at the top,diminishes the size of the lower line.
A
FASHION AND APPAREL DESIGN
4. Find optical illusions in clothing at home. , Draw examples of clothing
that will make the figure appear taller, shorter, heavier, and more
slender. (If you can't draw well, bring at ieast 10 pictures).
5. Find examples of clothing using dominant vertical or horizontal lines
that carry the eye across or up and down the design. (Find or draw at
least 10 pictures.)
Good proportion in spacing and a pleasing relationship of sizes and shapes
result in a design that is harmoniously related and unified.
Simplicity l otters! , mr.The location of the waistline divides the silhouette and creates a proportion
relationship.
6
FASHION AND APPAREL DESIGN
The principle of proportion is concerned chiefly with spaces, sizes, and shapes
within a design and their relationship to each other. To illustrate this
principle, examine the rectangles. In both A and B the division of space is
very obvious and therefore considered not as satisfying as C.
A B
In general, an uneven yet similar relationship is more interesting than either
an even or completely dissimilar one. To be pleasing and in good proportion,
the relationship should be neither too obvious nor too hard to see.
There are some laws of proportion in geometry that mathematically describe the
best possible relationship between areas. In the rectangles just shown, we
saw that the 1:1 ratio (with parts the same size) in A and the 1:2 ratio
(with the larger part twice as large as the smaller) in B are not pleasing
because they are too obvious. The pleasing ratio of rectangle C approximates
the 1 to 1.618 ratio. This is known as the "golden ratio.''
This golden ratio, I to 6. , enters the picture in designing width
and placement of borders, width of bands and their spacings, placement of
crosswise seams, size of pockets, and spacing of buttons.
7. Find a picture of some clothing that illustrates the golden ratio design.
Draw the rectangles on the picture to show this.
7
FASHION AND APPAREL DESIGN
8. Experiment with the placement of the waistline seam and the length of a
jacket on identical silhouettes. What effects do the placement of the
waistline and jacket length have on the figure?
9. Analyze several garments and classify them as good and poor design. (Get
pictures from magazines.) Give reasons for your decision.
This completes the few examples of geometry principles as applied in fashion
tailoring. However, geometry is just a start. One really interested in this
field needs good training in both sewing and art, for both of these comprise
the backbone of this career.
NAVIGATION
Navigation had always fascinated Dan. Maybe it was more the dream of sailing
the high seas or flying to far-away places. But for whatever reason we make
choices for our life's occupation, Dan wished to investigate navigation as a
career.
Dan's search for information took him to the school's counseling office where
he discovered that much of today's navigation is done by instruments with help
from computers. He learned, however, that all of the people responsible for
getting a ship or an airplane from one place to another must be able to figure
navigation routes "manually." In air transportation, most navigation is done
by the copilot and he probably received his training in a very specialized
school or in the armed forces.
Special licenses are required f6r most people involved in public transportation.
Tests are taken in order to show that one can properly do what is required.
New equipment means further training and the navigator, as well as pilot and
copilot, find advancement and pay increases often tied to their ability to do
the complete job well.
The great responsibility a navigator has necessitates that he does his
job well. This responsibility also allows him to command much as far as pay,
working conditions, and fringe benefits are concerned. Navigators enjoy most
of the benefits as copilots do. These include up to $30,000 a year in salary,
a flight schedule of 60 to 85 hours per month, and generally the finest work-
ing conditions. The burden of responsibility is great, however, and the need
to keep alert and well prepared at all times is required. During his investigation,
NAVIGATION
Dan found that in the early days of navigation a need for distinguishing one
point on the earth from another was needed. The earth, being two-dimensional
(height and width) required a different system of locating a point from
determining a one-dimensional line. With the line, all that is necessary is
to establish any one point, give it a number (usually zero), choose how long
one unit will be, and mark of all other points.
Example:
a. Draw a line
b. Establish a point
and how long one unit is
c. Mark off all other points as follows
I
-3 -2 -1 0 1 2 3
The "unit" of measure can be anything: feet, miles, degrees, etc.
The earth as a two-dimensional object will need two measuring lines, one for
height, another for width. We can show the ideas as follows:
With this idea in mind, Dan was shown the application-navigators use on the earth.
The measurement of height is done by lines of latitude; they enable
us to tell how many degrees north or south of the equator (the exact middle)
NAVIGATION
a given place is.
Notice that each line in the diagram is parallel
North pole90"
80° 0°10°
8/0GO' 60'
to the others, thus sometimes the lines of
20'
10'
0*
10'
60°40'
60'
30'40°
latitude are called parallels. To distinguishNorthern hemisphere 30°
20'
the 10° latitude above the equator from the 10°Equator
10'
0'
10'latitude below, latitudes above the equator are
called north latitudes, and latitudes below are
called south latitudes. Thus point B is at 40°
north latitude. But so are an infinite number
of other points (as point A).
The second dimension, width, is measured by lines
of longitude called meridians. Lines of longitude
are imaginary lines running from north to south.
Again notice that degree readings are symmetrical
with respect to the 0° meridians. (0° meridian
was chosen to go through Greenwich, England.) The
degree meridians to the left of the 0o
meridian is
a west longitude reading, those to the right are
east longitude reading; thus point C is 30° W.
3
20'Southern hemisphere 20°
30' 30'40' 40°
60'60' 60°5
0
080.
90'South pole
Parallels
North pole
Northern hemisphere
(50° 30 10° 0° 10° 30' 60°
k ii)Southern hemisphere
South pole
Meridians
NAVIGATION
Find the latitude and longitude of each of the points on the graph.90°
30
60°
8C 80°0°
Nliiir 60°
5.:iV
5o'
ive X 3o°
20° 20°
0° 10°
0 7e se 30° to° o' leOn EQUATOR
lilk
50°
OUTNERN HEMISPHERE ji;r410,3,
5o°
ids\ lb
6o 60°70
70°
,/-I-
NORTHE RN HEWSPFNMZ
t -4 -1
30 SO° 70
1, Point A
3. Point C
5. Point E
6. Analyze the diagram and find two points that appear to have more than one longitude
running to them
go°
go°
2. Point F
70"
4. Point D
90'o°
lo°
20°
We have stated that meridians or lines of longitude are drawn from the North
Pole to the South Pole. In order to guide the air navigator, a mariner's (or
magnetic) compass has been used. All directions on the compass are measured
from true north.
4
NAVIGATION
The no mark must be pointing to true north.
Then all that must be done is to align the
direction of the airplane through the center
of the compass to whatever direction it is going.
If the direction is through point Q,a reading
of 45° T("T" stands for true north) would be
taken.
Mariner's compass by 10 mark's
N
On the compass, plot the following readings by starting
at the center and drawing a dotted arrow through a point on the
circle. Label the points on the circle by problem number.
7. Due south
8. 270° T
9. Northwest
10. 110° T
Use the compass as reference to answer the following questions.
11. Find the reading for a course of due east.
12. Find the reading for a course running southeast.
S
Don developed a new appreciation for circular measuring after working with the
mariner's compass. One question that stuck in his mind was about the affect
wind had on the airplane's course. It seemed reasonable that with no wind all
a navigator would have to do is set a course at the correct angle from the point
of departure to the destination. He was soon told that the most difficult part
of the navigator's job is to make heading corrections caused by wind variables.
5
NAVIGATION
To see the effect that wind has on a plane, Dan was given some graph paper
and the following information:
Direction of the plane is 90° T.
Velocity of the plane is 160 mph.
Direction of the wind is 180° T.
Velocity of the wind is 60 mph.
Dan's investigation contained the following steps.
Step One: Starting at point A, Dan plotted the
original direction and velocity
(using one square = 20 mph).
Step Two: On the same diagram, he plotted the
wind's direction and velocity.
Step Three: He then formed a rectangle by
drawing in the dotted lines.
wind
4
wind
1
plane
I
1
20 140160 80 100 , 120 1140'1601
1----41/4 +-
, ., , , , ..1_ ..
20 40 60 80 100.120 NO 160
r 20{
40i
6011*
plane
i 20: 90 60 80--i0 10
1
wo 120 . No10 10 it
160
f---
20
-I-
1 160
ii---+-- N!
i !
.
-1t --r-
i i
NAVIGATION
Step Four: Dan could now find the resultant
ground speed and direction of the
plane by drawing the diagonal of
the rectangle from point A.
wind
I
40
160 i
plane
20 40 60 80 100 120 14 0 (60;0 e
_
11
I- I
1
ork
.1
resultant direction and velocityof plane
An easy but often inaccurate method of finding the resultant speed is to mark
off the length of the diagonal on paper and then measure it by using the graph paper.
The resultant direction is formed by measuring the angle formed by the diagonal.
Use a protractor to find this. (Ask your teacher for help if you do not have
or know how to use a protractor.)
How far off course has the wind blown the plane?
13.
14. The new direction of the plane, if not corrected for wind, would be
a. (Be sure to figure in the 90° T heading the plane started on.)
b. The resultant ground speed (AB) =
7
NAVIGATION
An airplane is heading due east. Its air speed is 200 mph. There is a wind
of 90 mph that has a direction of 1800 T. Find the resultant speed and dir-
ection of the plane (have each unit represent 20 mph).
.
4
H
4-
-L-
15. The resultant ground speed is
16. How far off couse has the wind blown the plane?
17. What is the resultant direction of the plane?
8
PAINTING AND PAPERHANGING
Bob's summer job as a painter's helper in his small home town was fine but he really
wanted to get into a larger community. He decided to check into becoming a painter
in some larger places. He wrote to the International Brotherhood of Painters and
Allied Trades, 1925 K St. NW, Washington, D.C. 20006. The information he got back
plus some checking with an area apprenticeship council gave him some answers.
He found that his apprenticeship and related training program would be much the
same as the other building trades. He discovered that the average hourly pay
for the building trades was $5.95. This didn't bother him because the pay varied
much from city to city. In fact, he found this table in the Occupational Outlook
Handbook.
City
Rate Per Hour
Painters Paperhangers
Atlanta $ 5.95 $ 6.20
Boston 6.08
Chicago 6.35 6.35
Cincinnati 6.23 6.85
Detroit 7.00 7.00
Houston 5.34 5.44
Newark 6.00
New Orleans 4.38 4.38
Philadelphia 5.22 5.34
Salt Lake City 4;87 5.07
San Diego 6.49 6.99
Spokane 6.17 6.17
PAINTING AND PAPERHANGING
After further investigation, Bob found that being a painter or paperhanger involves
more than just applying paint and paper. He would have to be able to mix paint,
match colors, and have a knowledge of color harmony.
Many painters and paperhangers work for contractors involved in new building construc-
tion. Many others are also employed by contractors to do repair and redecorating
work.
Painters and paperhangers may become foremen or advance to jobs as estimators for
painting and decorating contractors. In this capacity, they would need to know how
to compute material requirements and labor costs.
Some painters and paperhangers may establish their own businesses. This is what
Bob thought he would eventually like to have -- his own painting and paper hanging
business. The mathematics required was mostly involving area and volume. Here are
some examples of problems one might encounter in this trade.
1. The cross section of the roof of a barn has the shape of an isosceles trapezoid
surmounted by an isosceles triangle. The bases of the trapezoid are 50 feet
and 34 feet and its altitude is 14 feet. The altitude of the triangle is 7
feet. Find the area of the cross section. Area
C ross section
2
PAINTING AND PAPERHANGING
2. One gallon of paint covers 400 square feet on weather boarding. Find the number
of gallons necessary to paint the two ends of the barn in problem 1.
Gallons =
3. A church steeple has the shape of a regular hexagonal pyramid. A lateral
1edge of the pyramid is 38 feet and each side of the base is7.....feet. The average
2
covering capacity of buff paint on weather boarding is 400 square feet per gallon,
one coat. Find the number of square feet of the steeple and the number of gallons
necessary to give the steeple one coat of paint.
a. Square feet =
b. Gallons
Hint
= lateral height (altitude of triangular
face)
There are six congruent triangles. A =
1.e.b2
Find,/ by Pythagorean theorem:
(102 442 = 3822
a.
PAINTING AND PAPERHANGING
Another way to find the area of each triangle is to use
the formula A =,s(s-a)(s-b)(s-c) where s= 1.01-b+c)2
and a, b, and c are the three lengths of the sides of the
triangle. With this formula it is not necessary to know
the altitude of the triangle.
4. Find the number of gallons of paint needed to paint the sides and tops of three
gasoline storage tanks, each 12 feet in diameter and 16 feet high. Apply three
coats of paint. One gallon of paint covers 500 square feet with one coat.
a. .Gallons =
At $5 per gallon, find the cost of the paint. b. Cost
5. What will be the cost of painting the walls and ceiling of a room 12 feet 6 inches
by 16 feet and 8 feet 4 inches high at 80 cents per square yard. Cost =
PAINTING AND PAPERHANGING
6. The four side walls of a room 14'-6" wide and 17'-8" long are to be wallpapered.
The walls are 8 feet high. Allow 54 square feet deduction for doors and windows.
How many square feet of surface will be wallpapered? a. Square feet =
Find out how may square feet one roll of wallpaper will cover. b.
How many rolls are needed to do the job? c.
Find out the cost of vinyl wallpaper and figure out how much it would cost for
the wallpaper to cover these walls. d. Cost =
5
1
PAINTING AND PAPERHANGING
. Find the number of gallons of paint needed to paint the building pictured.
The length of the building is 38 feet and the width is 26 feet. The height
to the eaves is 18 feet. The roof is the gable type, the pitch being one
fourth. The building is to be given two coats. One gallon covers 400 square
feet with one coat.
Pitch = I means rise 1. or rise = span4 span 4 4
You can figure out the height of the gable end from the pitch.
The height of the gable end is feet-
6
PAINTING AND PAPERHANGING
8. A building having a semi-circular roof is 20 feet long, 12 feet wide, and 8
feet high to the eaves. Find the total area of the building (excluding the
0-0roof) to be painted. (A = I/ x R2. Userr = 3.14.)
Area
2 0'
8'
9. If three pints of thinner are used in 5 gallons of paint, how many gallons of
thinner are needed in 48 gallons of paint? Gallons =
10. A utility cabinet is to be painted inside and out-ide. It has four shelf spaces
(three shelves). If it is 50 inches by 24 inches by 12 inches, what is the
total surface in square feet to be painted? Disregard the thickness of the
material and shelves. (Shelves are 24" 3( 12".)
Area =
7
LANDSCAPE TECHNOLOGY
Jack thought that some day he would like to get involved in agriculture. He
decided to find out what was available for careers besides farming He was more
interested in land development and plants than in. livestock.
After a little investigating, Jack found there were three very closely related
jobs that interested him: 1. Landscape architect
2. Landscape gardener
3. Greenkeepers
These occupations involve landscaping and caring for grounds. The "landscape
architect" plans and designs land areas for all sorts of projects such as parks,
airports, highways, schools, and commercial and industrial sites. Jack found
that if he were such an architect he would prepare site plans, working drawings,
specifications, and cost estimates for land development. He could also supervise
the construction work based on his plans.
The "landscape gardener," Jack discovered, plans and works on small-scale land-
scaping and maintaing the grounds of homes and institutions. He might prepare
and grade the terrain, apply fertilizer, seed the lawn, or sod lawns. He lo-
cates where to plant shrubs, trees, flowers, and provides or supervises their
care.
The "greenkeepers" supervise and set up details for workers to keep a golfcourse
in good playing condition. Here Jack would determine work priorities and assign
workers to specific tasks, such as fertilizing, seeding, mowing, raking, and
spraying.
LANDSCAPE TECHNOLOGY
Jack was concerned with how much mathematics was needed to become a "landscape
technician" of some type. He went to a local vocational school and got some
sample problems. He also got himself a job at the Green Giant Garden Center.
hoping to be able to get some practical experience.
At the area vocational school, Jack discovered that this course of study takes
18 months to complete. He found out that some of the topics covered are: plant
materials, turf culture, landscape planting, sales, and management, and all
students spend three months on the job with their work supervised by both the
emplo-yer and the school. Jack was curious about what kind of work he would
be doing both during the on-the-job part of the course and after. He was told
that most technicians work for government agencies in the parks, golf courses,
and orchards that they operate. Some work for private individuals or corpora-
tions doing the same kinds of things. A growing number work for nurseries or
landscaping services.
It seemed to Jack that this was just what he would like to do and the job didn't
require the college training that a landscape architect would have to take. It
also would give him more of a chance to be outside, doing the actual work of
preparing and managing the landscape.
Let's look at some typical problems Jack could be requll.red to do.
1. You have been asked to fertilize the trapezoidal field pictured below by
applying three pounds of nitrogen per 1,000 square feet. Suppose fertilizer
costs $.45 a pound and the analysis is 24-16-12.
2
LANDSCAPE TECHNOLOGY
a. Find out what the analysis 24-16-12 means.
b. How many square feet is the field?
c. What will the fertilizer cost?
A = ih(a+b)2
A = area, h = height, a and b = lengths of the parallel sides.
200
o
2. The directions of a pesticide read three tablespoons per gallon of water. How
many cups would you put in 50 gallons of water? cups = .
Table
1 tablespoon = 3 teaspoons
1 ounce = 2 tablespoons
1 cup = 6 ounces
3. Use the picture on the next page.
a. How many tons of pebbles would be needed if it is spread 4" thick?
(Figure 1 ton = 1 cubic yard) tons =
b. How many feet of redwood 2" x 4" would be needed to outline the
pebble bed?
LANDSCAPE TECHNOLOGY
c. How many cubic yards of black dirt would be required if you spread
it 4" thick on the lot where grass is required?
d. How many pounds of seed would be required if you apply one pound
per 300 square feet?
House-yard Plan
4
LANDSCAPE TECHNOLOGY
4. a. Locate a tree so it will he equidistant from three roads: AB, BC,
and CA shown below. (Use a compass and straight edge. Show your
construction marks.)
b. Locate three shrubs so that they are equidistant from the tree and
each of the corners A, 13, and C. (Use a compass and straight edge.
Show your construction marks.)
5. Find the cost of sodding a lawn 132 feet wide and 152 feet long at $.56 per
square yard.
6. The formula for finding the area of an ellipse is A 41%b where a =2
the major axis and b is = 1 the minor axis. If you make a garden in the2
shape of an ellipse whose long diameter is 22 feetand whose short diameter
is 18feet, what is the area of the garden? Area =
5
LANDSCAPE TECHNOLOGY
7. Find the total area of a regular hexagonal walk surrounding a flower bed.
Area =
The walk is made up of six trapezoids. Since all six trapezoids have equal areas,
solve for the area of just one and multiply by six. A representation of each
trapezoid is shown here.
A1460° L'E
h
1;F- 60°
h
0' 6' C
A = .1,h(1) + b2) (formula for the area of a trapezoid)
8. Find the volume of concrete needed for the patio pictured, if the depth is 4".
(Hint: first find the area of the trapezoid, then the areas of the three
semicircles.) 28'
6
CARPENTER
Ever since Bill had watched some carpenters build a house next door, he thot;ght
it would be fun to build a house of his own some Jay. Maybe he would become
a carpenter.
Bill knew that a neighbor had been a carpenter at one time and went to see him.
While talking to the neighbor, Bill discovered that he was working at a local
technical school, so could tell about the schools, too.
The neighbor told Bill that carpentry is the largest of all of the building
trades, numbering over 830,000. He found that as the construction industry
grows, so will the need for carpenters. When Bill's neighbor left carpentry,
he was earning $6.60 an hour, but told Bill that wages run from $4.45 to $8.10
an hour depending upon the city. Most carpenters also receive some fringe ben-
efits such as insurance and pension funds.
The apprenticeship program for carpenters usually lasts four years. The appren-
tice must also attend a twelve-hour per month related training class. During
this time, the apprentice receives from 50 percent of journeyman's rate of pay
to 85 to 90 percent at the end of the four years.
Some prospective carpenters would go to the school where Bill's neighbor was
working. This school is a local area technical school (tuition free for high
school graduates under 21 years of age). At a school like this, everything
from blueprint reading to "stairways" would be studied. Other units would
include framing, tool care, safety, construction problems, human relations,
and communications. Courses at these schools usually run for one-and-one-
half years. Some of the apprenticeship time could be spent in one of these
trade schools.
CARPENTER
Bill was taking a course in geometry that year. He knew that his math courses
in the past could help him if he wanted to do carpentry work because of the
need to be able to compute with fractions and decimals, but how could the
course of geometry be used? He decided to explore the possible uses of geo-
metry in laying out and building a house.
While talking with the former carpenter now teaching at a local technical
vocational school, Bill was given some examples of how he could apply his pre-
sent course of geometry.
One of the first problems encountered is setting up the lines of excavation
for the foundation of a building. Here he found that the formula called the
"Pythagorean Theorem" was very useful. Bill knew that for any right triangle,
h2 = a2 + b2, where h represented the length of the longest side (hypotenuse)
and a and b represented lengths of the shortest two sides (the altitude and base).
2
altitude
CARPENTER
Now if h2 = a2 b2
then h =1/7777
b =1/h- - aL
a --417
1. In the figure below; rectangle ABCD represents the lines of excavation
for the foundation of a house. Figure the lengths of the diagonals AC
and BD if the side AB is 49'-0" and AD is 36'-0".
F
E / Building lines
H
Diagonals are and
2. In checking '_he corner D to find if it is square, GD is laid off equal to
3'-0" and DH is laid off equal to 4'-0". What should the length of the
segment CH be?
3. To square up corner A, we lay off AE equal to 6'-0" and AF equal to 8'.0".
What should we then make the line EF?
4. Rafters had to be orchred for the building. Bill was asked to figure out
what length rafters he should purchase for the building. In the illustra-
tion on the next page what is the line length of the common rafter of the
gable roof shown if it has a span of 36'-0" and a rise of 9'-0"?
3
CARPENTER
4. Line length =
SPAN
5. If you want an overhang of what length of lumber must be purchased
to make common rafters for the building? Rafter =
6. It was necessary to order sheeting for the walls and gables of the building.
Using the dimensions of problem 4 and knowing the height of the walls
to be 8'-6", find the area of one gable end of the building.
Area =
What is the least number of 4' x 8' sheets of plywood it would take to cover
one gable end. Number of sheets .,----
Bill asked the vocational instructer if he had any material on carpentry that
he might study to find applications of geometry on his own. On the next pages
are some of the situations Bill came up with.
4
CARPENTER
7. Figure the number of square feet of form
work required for the end of the concrete
retaining wall shown here (find the area ofthe trapezoid).Sq. ft.
11.-3"
8. A carpenter is to construct a gable-roofed shed. The section has the
shape of a triangle with a base 24 feet wide and base angles of 30" and
45° respectively. Let 1 inch equal 12feet and construct this section.
How long are the rafters? and
9. Figuring 100 square feet to a "square" of asphalt shingles, how many squares of
asphalt shingles are needed to cover the roof shown here (make no allowance
for waste)?
10. The basement of a building 30 feet long and 24 feet wide is to be ex-
cavated to a depth of 6feet Find the number of cubic yards of material to
be removed.
5
1
CARPENTER
11. A basement foundation is shown below. One corner is removed showing a
cross section through the wall. The broad base upon which the wall rests
is the footing. The foundation is 30 feet long, 24 feet wide. The wall is I
foot thick and 7'-6" high. The footing is 2 feet wide and 1 foot thick.
Find the number of cubic feet in the walls and footings.
Volume =
6
DRAFTING AND ARCHITECTURE
"Draftsmen translate the ideas, rough sketches, specifications, and calculations of
engineers, architects and designers into working plans used in making a product."
The Occupational Outlook Handbook tells of the 310,000 draftsmen working mainly
for private industry. Most of these people have attended a vocational school or
junior college to receive their training but a few have acquired their skills
strictly through on-the-job training programs.
Most draftsmen start out as tracers earning approximately $6,000 a year. They
look to a good future because design and the problems connected with it are be-
coming m. re important as the years go by. They will need to upgrade their skills
1wever, because modern means for producing drawings and new electronic equipment
will eliminate some of the lower level tasks now done by some draftsmen.
The architect differs from the draftsman in numerous ways. Although some architects
find themselves often doing work similar to that of the draftsmen, most of their
work usually deals with the idea behind and the design of an item. They most often
have had five or six years of college and have passed the requirements for licensing
that the individual state has established.
An architect needs an interest and ability in technical problems and their sol-
utions but in addition he needs to have interest and ability to sense and see
artistic concepts well. These two areas do not always fit together in one individ-
ual and should cause the would-be architect to spend some extra time looking at his
abilities and interests.
Because more training is required to become an architect, architects usually command
more pay than a draftsman does. Most begin at $8,510 or $10,528 if they go to
work for the federal government. Their working conditions are good and chance for
advancement is great if they possess the qualities desired by employers.
DRAFTING AN!) ARCHITECTURE
Draftsmen and architects are skilled technicians who work at a drawing board.
Their jobs are to prepare working drawings of buildings, parts of buildings, or
equipment for buildings. They take a designer's imagination and transform it into
a builder's reality. They correlate the architectural, structural, mechanicai
electrical, and shop drawings with the building contractor's needs. To become a
skilled draftsman or architect requires some background in mathematics. Let's
take a look at some mathematics you will be able to apply in these two careers.
1. Construct a triangle having a perimeter of 8 inches and the ratio of sides
2:4:5. (Hint: See unit "Standard Constructions," problem 9.)
2. The hypotenuse of a ri,ffit triangle is 4" and one angle is 30°. Construct the
triangle.
2
DRAFTING AND ARCHITECTURE
3. To reduce a polygon so that the given sides become proportionately smaller
is a common problem since most drawings are scale drawings. Reduce the3
polygon ABCXDE to a polygon so that the given side AB becomes 4 inch.D
Follow these directions:
a. Draw a given polygon ABCXDE.
b. Draw a segment A'B' parallel to AB so that the length of A'B' is 4 inch.
(A'B' can be drawn anyplace. It would be easiest to put it to the right of
the picture. Just be sure it is parallel to AB.)
c. Draw AA' and BB' and extend these segments until they meet at a point
we'll name O.
d. Join the points C, X, D, and E to O.
e. Through B' draw a line parallel to BC to meet CO at the point we'll call'C'.
f. Through C' draw a line parallel to CX to meet XO at a point we'll call X'.
g. In the same way find points D' and E'.
h. Now A'B'C'X'D'E' is the required polygon similar to ABCXDE.
In order to convey their ideas, architects and draftsmen must be able to fulfill
two main requirements:
I. To give a clear and accurate picture of the shape of the object.
2. To show the size of the object.
3
DRAFTING AND ARCHITECTURE
The drawings are a system of views showing separate drawings which give details
in accurate proportion of the different sides and faces of the object. The
object has three dimensions, but when drawn on a sheet of paper any one view
shows two dimensions only. At least two projected views will normally be required
to explain fully the shape and size of an object. The diagram below shows
projected views from the front, back, and both sides as seen by the eye. Only the
edges or outline of the object are shown as full dark lines; those edges that are
not visible from any particular view are shown by dotted lines. The method you
see illustrated is called "orthographic projection," and by using it many "projected
views" of an object can be drawn to exact scale.
4
DRAFTING AND ARCHITECTURE
4. Draw full size the :rout elevation (vie), end elevation (view), and plan
of the details shown in the object below. (Use the example just discussed
as a model of what to do.) WritL. the dimensions in :teach drawing.
6
DRAFTING AND ARCHITECTURE
5. A draftsman is required to draw a circle for a pulley 2'-9" in circum-
ference. What radius should he use? (Useir = 3.14).
Radius =
6, A special electronic device requires that 102 holes be equally spaced on a
circle whose diameter is 2.75".
a. What is the central angle between the holes?
b What is the distance on the circumference between holes to four decimal
places? (Hint: Make calculations from center to
center of the holes.)
(Hint: See unit "Circle Relationships" about length of an arc.)
7
DRAFTING AND ARCHITECTURE
7. A rectangular building x 135' is drawn to a scale of 1" = l'-0".8
What is the size of the rectangle that will represent this building on a
blueprint? Size =
8. The wing span of a jet plane is 56'. What will be the wing span of a model
of this plane built to a scale of I" = 8'-0" Span =
9. Draw a plan to a scale of 1" = 1'-0" to represent a room4
20'-0" long
by 16'-9" wide.
OPTICAL TECHNICIAN
"Over half the ropulation of the United States over the age of six wears eye.-
glasses." This MAV be surprising but our concern Vor good eyesight has made
it true. It is becoming increasingly important for us It) find quality eyewor.
Afto:- an examination has determined that glasses are necessary and
prescription has been written, two people have a great deal to do with the glasses.
The first person is usually called the optical mechanic. He grinds the glass so
that it will do what it must for the oyes. He also fits the lenses into the
chosu frames. A dispensing optician then fits the eyeglasses to the face.
Often an optical technician will perform both these duties.
Throughout the country, most people learn their trade on I the job. Recently,
however, many trade schools offer programs. Training is available at both Anoka
and Eveleth. These programs last for a little loss than a year and include such
courses as grinding, theory, lens hardening, anatomy of the eye, and frame repair.
Nany prospective technicians at the area trade schools never complete the course.
There is such a demand for trained optical technicians that: jobs are offered
before the whole program is completed. When they do begin work, technicians
earn a wide range of wages, depending upon skill, responsibilities, and location;
usually, however,hourly wages run from $2.50to $4.25 a hour. Purchasing one's
0;,,n business opens up greater possibilities for increased income.
The mathematics required On the job is rather et;tensive. The use of tables
makes the job much easier, but a thorough underWinding of certain concepts of
geometry and algebra are essential. It is difficult to extract the math used
by the optical technicians and put it in meaningful career-related setting.
OPTICAL TECHNICIAN
There is much background information needed to lead up to the mathematics
applications. The geometric concepts covered in the remainder of this unit
are typical of those used in preparing for the job and on the job.
Light and its reaction to glass must be studied and understood. When a light
hits a mirror, the angle of reflection is equal to the angle of incidence. The
angle of incidence is formed by two rays: first, the ray of light hitting the
mirror, and second, a ray perpendicular to the object the light is hitting drawn
from the point where the light hits. See figure 1.
fig. 1
1. So if 4:1 = 35°, the angle
of reflection,L r, is
2. If we wished to bounce a light
fig. 1
Normal
4To Mirror
-
Light Source MirrorBelle( hom .1 plant. mil rm.
< i :Ingle of in( r (,1
reltectIo11.
L. I'
Li
fig. 2"around" a corner (see figure 2)
A
to point.A from any point B, what
would the angle of incidence
be? 1111111 1111111 1111112V[------
Well
Draw point B where this would occur.
Mirror
OPTICAL TECHNICIAN
Fig. 3 shows that the reflected ray is the
same as the incident ray on convex and concave
mirrors. The light ray hits at the center, or
apex of the curve in these examples. So, even
though the mirror is curved, the light will be
reflected a+- the same angle it comes in.
?ef/.--:cr(f'd R
fig. 3
Light SourceCoNvrx mirsRotz
11,111 .1 CUM ("N.
Ref 1
NovmdI
CONCAVE MIRROR
ktk111.)11 IhM1 (1111CO\
The question then arises,"But what if the light does not hit at the exact
center? Surely it will then be affected by the curvature of the mirror." The
answer is, yes it will. But the effect is a geometric arc; therefore we can
predict what course the light will take after hitting the mirror. Figure 4
shows the geometric relationship betweeft incoming light and reflected light on
a convex mirror. Let's analyze how A is formed Light source emits a ray
that hits at point X, The light ray is parallel to the NORMAL LINE. At a point
called the virtual focus (a point midway between the'surface of the mirror and
the center of its curvature), B is found by drawing a line from the virtual
focus to point X.
3. Is m B = m A?
If so why?
3
fig. 4CONVEX MIRROR
Light Souvce 1.ptx
0 -4Vir+uol Focus
Normal Linec'
/---Radiuspty
Light Source 2. / z#
OPTICAL TECHNICIAN
p Notice that A makes the reflected ray diverge (go outward). Light source 2
hits at point Y. The D is formed by a ray from the virtual focus through D.
4. If ml. C = 60°, at what angle does light 'source 2 reflect from the
mirror?
Figure 5 shows the reaction of light as it hits a concave mirror at a point
other than the center. Notice the difference in the paths of reflected light
on the concave mirror as opposed to the convex mirror.
5. The convex had divergent rays that went outward, the concave mirror's
reflected rays go
fig. 5
CONCAVE MIRROR
and are convergent at point A.
Point B on figure 6 is the point of convergence.
fig. 6
. If a light source stems from a point A parallel to the normal line, what
is the angle of reflection formed at the mirror's surface?
(Draw the angle in figure 6; call the point where the light hits the mirror\
point D.)
4
OPTICAL TECHNICIAN
7, Locate point C below the normal line whose reflection would form this
same angle to point B.
An instrument used extensively in the layout and checking of lenses is the pro-
tractor shown below.
Reference points arc the horizontal and
vertical axes. A lens is laid out so as
to obtain the vertical and horizontal
placement of the optical ce-ter. As
you can see, this protractor works
on the same principle as the one
used in construction of ordinary
fig. 7
90v....:901111.4./a/T
-uuselopsime*
10 5 l :It 5 10
geometric figures; therefore, any- \\.---
111:
.-',
.Y- /.
,..------ ,
,one interested in the optical tech- 0,.
/,.
s
nician field would have a gOod back-
ground i.f he knew how to use the pro-
tractor used in geometry class.
/1,-r'."
90
44"
In optics, prisms are used to diSplace or change the direction of light coming
into them. Certain types of light can be taken out of incoming rays.
8. With reference to fig. 8, list
the faces of the prism.
5
OPTICAL TECHNICIAN
9. List the base of the prism (all light is refacted towards the base of a
prism) A base is defined to be any
plane perpendicular to the base-apex plane (plahe ABCD in figure 8).
10. Plot and label (using IJKL as points) another base (other than ORTZ) for the prism.
As you can see, an optician must know the qualities of angles, a protractor, and
prisms as well as right triangle relationship and many other geometric properties
we have omitted.
SHEET METAL
Some geometry students were considering careers as sheet metal workers. They
were concerned about the good they would get out of a course like geometry.
Could they ever use it later? A sheet metal instructor was invited to the
geometry class to discuss this problem.
He presented information of which most of the students were not aware:
"Sheet metal workers usually need to complete a four-year
apprenticeship program before they can become journeymen.
"This apprenticeship training includes on-the-job and class-
room training.
"An apprentice receives between 40 and 80 percent of journey-
man pay.
"Average pay received by sheet metal worker in 1970 was $6.75
per hour.
"As construction rates remain high, so will the need for sheet
metal workers.
"Sheet metal workers need to be able to picture, in their
mind, what the finished article will look like (spacial aptitude).
"Almost all sheet metal workers in the cities are union members.
"The vocational-technical schools, of Duluth, St. Cloud,
Minneapolis, and St. Paul have programs in sheet metal that
last from 18 to 21 months."
In addition to hearing the sheet metal instructor, a few students had their math
teacher help them set up a tour of a local sheet metal shop. They went primarily
to find out what kinds of geometry are needed on this job, but in addition
they discovered that sheet metal workers need other abilities. For instance,
SHEET METAL
much time was spent reading blueprints and preparing layouts so the flat sheets of
cut material could be turned into three-dimensional objects. Here is where the
ability to visualize a finished product became important.
It didn't take long for the geometry students to see that these workers need to work
with some pretty interesting math concepts. Here are some typical problems a sheet
metal worker might need to solve.
1. A cylindrical smoke stack made by bending rectangular sheets of tin is to
have an inside diameter of 12 inches and a height of 10 feet. What is the
surface area of the smokestack? (The formula for the lateral area of a
circular cylinder is A = 21Yrh where h is the height.) How much material
will.be necessary if the sheets of tin are 14 inches wide by 10 feet in length
and if two inches are allowed for overlap ? a.
b. (Hint: Use 7 = 3.14J Round up any fractional part of a sheet.
2. You are required to make a tank within the following limits: it must be six
feet high and contain 400 gallons of water (1 gallon = 231 cubic inches). What
must the diameter of the tank be? DiaTeter = a.
How many square feet of sheet metal are required to make this tank, assuming
ot.no overlap? (Remember the formula for area of a circle is A = h r
2.)
b. square feet (Use n-= 3.14) (It would be helpful to use a caculator.)
SHEET METAL
3. What is the capacity of a tank 10 feet long having an inside diameter of
three feet? Capacity = a.
What are the dimensions of a cubical tank of the same capacity (1 =
Use the cube root table attached. b.
O"
X \
In the same way that a square root can be represented as the side of a square,
a cube root is the side of a cube. The symbol for cube root is \i"--7 To find
the cube root of a number, we use the table shown. You will notice from the
table that there are only five perfect cubes
Perfect cube
from I
Cube root
1 =
to 149. These are:
(3/---)
8 2 =
27 3 = 27
64 4 = 31,167i
125 5 =3v 125
liW, Cs;ht- MI Abulit L. 2 INCHESt P,t is ',ICA II V.ON1AINS
H CUHIL INCHES
2
SHEET METAL
Table of Cube Roots (1 to 149)
0
1
2
3
4
5
6
7
8
0//
12
13
14
0 1 3 4 5 6
1 1.26 1.44 1.59 1.71 1.82
2.15 2.22 2.29 2.35 2.41 2.47 2.52
2.71 2.76 2.80 2.84 2.88 2.92 2.96.
3.11 3.14 3.17 3.21 3.24 3.27 3.30
3.42 3.45 3.48 :3.50 3.5:3 3.56 3.58
3.68 :3.71 3.73 3.76 3.78 3.80 3.83
3.91 3.94 3.96 3.98 4 4.02 4.04
4.12 4.14 4.16 4.18 4.20 4.22 .4.24
4.31 4.33 4.34 4.36 4.38 4.40 4.41
4.48 4.50 4.51 4.53 4.55 4.56 4.58
4.64 4.66 4.67 4.69 4.70 4.72 4.73
4.79 4.81 4.82 4.83 4.85 4.86 4.88
4.93 4.95 4.96 4.97 4.99 5 5.02
5.07 5.08 5.09 5.10 5.12 5.13 5.14
5.19 5.20 5.22 5.23 5.24 5.25 5.27
01.91
2.57
3
3.33
3.61
3.85
4.06
4.25
4.43
4.59
4.89
5.03
5.16
5.28
8 9
2 2.08
2.62 2.67
3.04 3.07
3.36 3.39
3.63 3.66
3.87 3.89
4.08 4.10
4.27 4.29
4.45 4.46
4.61 4.63
4.76 4.78
4.9(1 4.92
5.04 5.05
5.17 5.18
5.29 5.30
4'754tLeft column
Top line 1-1. Value in table
If the number is between I and 149, but does not appear in the table," approximate the
answer as shown in the following examples.
EXAMPLES:
1. Find V 121.7 to one decimal place.
p,-----The table shows that e/-1-17. = 4.95 and e/7-27.= 4.96. Thus V 121.7 is nearly
4.96, since 121.7 is closer to 122 than it is to 121. The answer is ?/121.7 =
4.96 approximately. (Note: Approximations made by use of this table will be
less accurate for cube roots of numbers from 1 to 20 than for numbers. from
21 to 149.)
2, Find the side of a cube whose volume is equal to 142.35 cubic inches.side = li\17775".
Using the table, we find that `,.3./1417 = 5.23 and 3 1V7272 = 5.22'
Therefore, S = Y1.72737 = 5.22", approximately.
4
SHEET METAL
4. How many square feet of sheet tin are required to cover the following
shape?
0
2'6
5. A sheet metal worker makes a form in the shape of a right triangle. The0
larger acute angle is 40 more than the smaller angle. Find each of the
acute angles. (Either use the fact tha'... "the sum of the angles equals0
180 ", or that "the two acute angles of a right triangle are complementary.")
The measures of the angles are and
6. Find the length of an arc in a piece of sheet metal bent to a radius of
3" at an angle of 90°. See figure below.
4A
°94Y1
angle of ao
radius of R
0The angle of 90 is the central angle.
a
length AB = 370 x 2/TR where the circumference of the cirle with radius
R is C = 211R. Length of arc =
5
SHEET METAL
7. A sheet metal worker is asked to cut a piece of copper in the shape shown.
When he starts to make his layout, he finds that he needs more information.
What is the least amount of additional information he needs to make the
layout?
8. On a circular piece of sheet metal lay off six equally spaced 1/4" screw3
holes with centers 4 " from the circumference of a 5" diameter circle.
See sketch below. (Draw a circle on paper and do this)
6
SHEET METAL
Many manufactured articles are produced in sheet metal from pressings and bln7*.-
ings, which are then joined at the seams or edges to form the completed compor
To estimate the cost, it is important to be able to find the exact shape of the sur-
face area of the complete part. To do this the object is "developed", i.e.
it is drawn unfolded along the joints and "fold" lines, giving the exact shape
of the area of metal sheet required to make it. To indicate the drawing con-
struction it is usual to mark the given views (called orthographic views) at var-
ious points (using letters or numerals), so that they may be properly related
to thendevelopment drawing." The following shows the development of a sheet
metal bracket and the frustrum of a square based pyramid. Remember that orly
true lengths can be used for the development, and be sure that these have been
derived correctly from the given views.
Sheet Metal Bracket
SHEET METAL
Frustrom of a square based pyramid.
SIMPLE
DEVELOPMENT
Allowance for folding, overlapfor seams and joints etc, notshown.
Joint line should alwiqs betaken on,the shortest length.Developments must use truelength lines notprojected"lengths.
9. Draw a development of the frustrum of a hexagonal prism shown below. To
do this the true length of each face of the hexagon is shown in the lower
plan (distance 1 to 2 for example). The hexagon is made up of six panels
which may be developed alongside the elevation (top plan), the height of
the prism should be transferred by direct projection of lines. The base
and the uncit portion of the top are then added to complete the develop-
ment layout.
8
ENGINEERING
Engineering, in 1970, had nearly 1,100,000 people employed, making it the second
largest professional occupation. Teaching is the only profession with more em-
ployed than engineering. Most of these engil .ers work in some area of the manufac-
turing industry; the others work in governmental agencies and schools.
-Usually, an engineer is required to attend school for 'four-to five years. This
time is spent studying both the basic sciences and the social sciences with the last
half of the prbgram emphasizing the student's engineering specialty.
licensed by taking and passing a rigorous examination.
He becomes
Some engineers have recently found themselves out of work. This has beent.,due, in
part, to the fact that defense spending dictates some of the needs for engineers in
manufacturing. If government spending changes or drops off, the employment of many
might be affected. In addition; our economic conditions will affect the need for
engineers.
In 1969-1970 a first-year engineering graduate with no experience earned $10,400.
If well-prepared and doing a good.. job, engineers earnings can increase rapidly.
Present engineers with 21 to 23 years' of experience earn an average salary of
$18,350.
A few of the larger branches of engineering are:
1. Electrical engineer
2. :Civil engineer
3. Mechanical engineer
4. Aerospace engineer
ENGINEERING
5. Chemical engineer
6. Metallurgical engineer.
7. Ceramic engineer
8. Agricultural engineer
Most engineers specialize in one of the may branches of the profession. Nevertheless,
the basic knowledge required for all areas of engineering often makes it possible for
an engineer to shift from one field to another. An engineer might encounter the
following problems:
The area of a sector of a circle is equal to e. fraction of the area of the
circle determined by dividing the size of the angle by 360.
A
Area of sector AOB =360-
11,K. r2
Sector AOB of X° in a circle
of radius r.
The area of a segment ABD (the shaded portion of the figure below) equals
the area of sector AOB minus the area of triangle AOB.
Find the area of AOB by the formula A =16(s-a) (s-b)(s-c) where a, b,
and c are ,lengths of the sides and s = 1 (a+b+c). If you know the2
altitude of A AOB, then use the forirula A = 1 *b*h. In some cases, you2
might have to find the area of 4-AOB using trigonometry.
2
ENGINEERING
Consider the following situations:
r
1. A gothic arch is formed by two arcs, each one sixth of a circle. The center
of each circle is at the ends of the width of the arch. The radii of the1
two arc sides eqdals each other and the width of the arch. (See the diagram.)
Find the area of the gothic arch of radius 6 ft.
Area =
2. -Running down the-center of a four -lane highway is a safety guard made of
concrete. The base of the cross-section of the guard is a trapezoid and the
top is a semicircle. The lower base of the tiapezoid is 18 inches; the upper
base (which is also the diameter of the semicircle) is 6 inches; and the
distance between the bases is 12 inches.
a.
b.
Draw a sketch of the cross-section showing the dimensions.t
How many cubic yards are there in one mile of the safety guard?
Cubic yards =
ENGINEERING
3. A solid cylindrical pillar of iron supporting a building is 14 feet high
and has a diameter of 9 inches. If it weighs 450 pounds per cubic foot, its
weight is .-about pounds. (Use 3.14 = ft.)
4. A field has the shape of the irregular polygon shown in the figure. AM =
45 rods; PC = 35.5 rods; OD = 64 .gods; RE = 30 rods; AQ = 6 rods;
AR = 30 rods; angle FCC = 30°. Find the area of this field in square yards.
(1 rod = 161/2 feet).
N
Area = square yards.
5. An officer of engineers, mapping a piece of land lying between a straight
road and a curved stream, obtains the data given in the table on the following page.
This table shcws that at various points from one end-of the road the distance
from the road to the stream is s yards and the distance from the point on the
road to the end of the road is d yards.
a. Using as a scale 1/8" = 1 yard, draw a map of the land. (Be sure to read
the following hint before attempting.)
b. Find its approximate area.
4
ENGINEERING
s 0 17 56 109 174 200 184 132 96 60 0
d 0 10 20 30 40 50 60 70 80 90 100
I
Hint: In actual practice, a large number of surfaces are enclosed by curves,
for; example the field AFKD, bounded on one side by a river :.
Fig. A
To find its area, first subdivide the field into convenient sections as shown
by the lines CB and RE. The area of the plane figure bounded by a curve
such as ABCD can be approximated from an accurate drawing of the figure by find-
ing the average height of the curve and multiplying this height by the base.
By the "average YAgl,t of the curve" we mean the height of a rectangle equal
in are to the area of the plane figure with the same lase.
M
0
If you consider the section ABCD of figure A above, PQ is the average height of the
N
3
Fig. B
curve from base AB.
ENGINEERING
The average height may be approximated by drawing the horizontal line MPN so
that the area DMP, which is added to form the rectangle ABNM, is as closely as
possible equal to the area CNP which is cut off to. form the rectangle. Since
the area added.i2 approximately equal to the area cut off, the areas of the
rectangle ABNM and of the plane figure ABCPD can be considered equal (for
all practical purposes)
a, 1" = 1 yard b. Approximate area =8
6
ENGINEERING
6. The dome of a building is in the shape of a semicircle that spans 60'.
What is the length of the vertical support 10; to the right of center.
Vertical support = feet.
60.-- o"
Remember that the length of the altitude of a right triangle, h, is the
mean proportional between the two segments, x and y, that it divides the
hypotenuse into.
7
x h
h y
ENGINEERING
7. Suppose that two straight railroad tracks (1) and (2) intersecting at a
point M, are to be connected by a circular track. The circular track is
to be placed tangent to tracks (1) and (2). Assume that it is a distant,
of 1000 feet from the point of tangency to the point of intersection
of the two straight tracks. Also, suppose the radius of the circular track
must be 500 feet. How far will the intersection of the two straight tracks
be from the circular track?
(In other words, find the length of PR.)
b. What is the relation between OT and TP?
8. How many cubic yards of soil will be required to fill a section of land
75 feet wide by 125 feet deep if the surface is to be raised. four feet in the
rear and gradually sloped to the front, where it is to be one foot deep?
73.
4'
ENGINEERING
9. In building a bridge, panels as pictured below are needed. In the figure,
the timbers CB, DF, and EG are perpendicular to CA. From the given dimen-
sions find the lengths of the timbers.
a. DF =
b, EG =
c. CF =
d. DG = 12'
e, AB =
B
Hint: Use proportions from corresponding sides of similar triangles.
9
MACHINIST
Alex heard that the counselor at his school would go over his school record with
him he mad.,1 an appointment. While talking with him, Alex thought about how
he could use all of this information about himself. The counselor thought this
information should be used as Alex planned his future.
After the interview, Alex decided to put down on paper some of the things that
he had leailt:ed about himself. He listei these items.
My grades are best in shop courses.
My interest lies in the mechanical area.
An aptitude test shows I have highest ability in spatial and mechanical areas.
My favorite hobby has always involved the building of things.
I once won an award for a project I made in machine shop ...
"Machine shop! That was my favorite class. Maybe a career in the machining
trades would be fun." Alex went back to the counseling office the next day to
look up some information about this area.
In the Occupational Outlook Handbook,-Alex found there are at least five specific
jobs in the larger category of machining occupations. They are: machinists, tool
and makers, instrument makers, machine took operators, and setup men. Each of
these jobs requires certain specific skills that make them different from one another,
but there are many similarities, too. Al? of these jobs usually involv:!,operating
power-driven machinery; they require accuracy and that accuracy involves precision
handwork.
The handbook continued, "All-round machinists are skilled workers who can operate
most types of machine tools. Machine tool operators commonly operate only one kind
of machine tool. Tool and die makers specialize in making dies for use with presses and
MACHINIST
diecasting machines, deVites to guide drills into metal, and special gauges to
determine whether the work meets specified tolerances. Instrument makers use
machine tools to produce highly accurate instrument parts made of metal or
other materials. Setup men adjust machine so that semiskilled machine
tool operators can run the machines."
Alex found that the'most comllon way to enter these trades is through an appren-
ticiship program of on-the-job training. For the machinist and tool and die
maker, this training can take four or five years.
There was much information in the guidance office about these trades but one
thing especially stuck in Alex's mind. It seemed that the future of
these trades could depend much on how computerized or numerically controlled
machines would Le used. At least the machinist will be doing different things
and will have to continue to improve his skills as the years go by. Machines
that are computerized can do the entire job of turning out a product -- if
instructed properly and maintained adequately. The people working with these
highly complicated machines will indeed need special skills.
Alex found that uses of geometry in the machinist trade are an everyday occur-
ance. Angle relationships, circle relationships,and polygons are the most fre-
quently used topics. Some of these relationships are shown below. For more--
information on these relationships, refer to the "Polygon" and "Circle Relation-
ships" units.
2
MACHINIST
Polygons
3 sides --- triangle
4 sides --- quadrilateral
5 sides --- pentagon
6 sides --- hexagon
7 sides --- heptagon
8 sides --- octagon
9 sides --- nonagon
10 sides --- decagon
3
Supplementary angles(Sum of 180°)
Complementary angles\(Sum of 90°)
Vertical anglesare congruent
'MACHINIST
The next semester Alex decided tc take an advanced machine shop class. One
of the first projects was to TI:achine, fr,--1 a piece of round stock, a "hex"
headed bolt. He knew that "hex" was short for hexagonal, meaning six-sided.
He decided to take the 2" (2" diameter) and lay out the six sides on one end.
\
After he cut the piece to the length he desired, Alex put it in a vise, end
up. Before gOing any further he made a drawing of what he wanted to do.
In order to get a six-sided figv..re in the
circle, he had to divide it into equal
parts. How may degrees would each part
contain? 1.
Gote: There are 360° in the circle]
Alex knew that the central angle (an angle whose vertex is at the center of the
circle) is equal in measure to its intercepted arc. Show how he inscribed a hexagon
in the circle provided above. After that, he ground the edges and cut the threads.
His next problem was to machine a'VL" (across the flats) square nut. In order9
to choose the correct round stock, he must find the length across the corners
(diameter). Use the diagram of the square nut to find the "across corners"
dimension. (Hint: c2
= a 2 + b2 )
2. Across corners dimension:
3
Across-Flexts 1,
1" = --' scale2
MACHINIST
3. What is the distance between centers of the
four holes equally spaced on a 6" circle?
X=
For the next problems, you will need the given sine table at the end of the unit.
4. A right triangular prism (fig.l) must be machined out of
round stock. Find the diameter of the stock needed;
use fig. 2 (drawn actual size) to do your work.
a. Circumscribe a circle around A. ABC
(Construct perpendicular bisectors of
any two sides. Where they meet is the
center of the circumscribed circle.)
5
TRIANGULAR
PRISM
MACHINIST
b. The diameter of this circle can be found by using the formula!
D=sin L- B
that is, diameter is equal v.Lo the length of segment b divided by the
SINE of angle B. (Sine of an angle is the ration of the
length of opposite side
Angle Sins
.0175,0349.0523.0698 '
.0871
.1045
.1219.1392.1564
Angle
46'47'48'49'500
51052053°54°
51(10
71').s.7114.7431.7547.7660
.7771.7880.7986.8090
length of hypotenuse
Use the sine tables at the right to find the sine
of B.
a. sin B =
1'
5°
6°7°
9°
b. Find the diameter of the stock needed. (Find D)
Diameter =
10° .1736 550 .8192
11° .1908 56' .8700
12° .2079 57°13° .2250 580 b14° .2419 59° .857215° 60° 8660
16° .2756 6i° .874617° .2924 62' .882918° .3090 63° .891019° .3256 64° .898820° .3420 650 .9063
21° .3584 66° .913522° .3746 67° .920523° .3907 68° .927224° .4067 69° .933625° .4226 70° .9397
26° .4384 71° .945527° .4540 72° .951128° .4695 73° .956329° .4848 74° .961330° .5000 750 .9659310 .5150 76° .970332° .5299 77° .974433° .5446 780 .978134° .5592 79° .981635° .5736 80° .9848
36° .5878 810 .987737° .6018 820 .9903380 .6157 830 .992539° .6293 84°40° .6428 85°
41° .6561 86° .997642° .6691 870 .998643° .6820 880 .999444' .6947 890. .999845° .7071
MACHINIST
5. Eight bolt holes are equally spaced on the
10" circle. What is the center-to-center
distance of any two adjacent holes? (Hint:
Kow large is a central angle? Draw appropriate
perpendiculars.)
Step i Find the measure of a central angle.
Step 2. X° = measure of a central angle.
Step 3. Look up the sine of Xo=(approximate 1o ),
substitute into the formula. Sin X° = a5
Step 4. Solve for a
a =
Step 5. Double a to get distance between
centers of two adjacent holes.
2a
7
MACHINIST
6. Another application of the sine function is to find the
missing length of a side of a triangle gi,,en an angle
and tho length of the opposite side.
Find length A,
Hint: Sin'. 45° = 1A
CEMENT WORKER
A cement worker or cement mason in often also called a concrete finisher. The
name concrete finisher is probably the best one to use because it most clearly
tells what work is done. In this area of work it is easy to see how specialization
has set in. On large construction projects, the mason that does the whole job of
putting in the concrete is gone. No longer does one man build form, mix and pour
the concrete, and then finish it. On most projects, the concrete finisher may
supervise some of the initial phases of the laying of concrete, but his main concern
is the last steps which make the concrete look like it does when it hardens.
Even this phase has various specialists, however. There are finishers that work
primarily with large power equipment on big projects (highways, etc.) and there
are others that work with mastic (a fine asphalt mixture that is applied over the
concrete).
A large number of concrete workers have obtained their training on the job. These are
people who work their way up as they gain experience and are able to do the job.
This trade does, however, have a three-year apprenticeship program requiring the
related classroom instruction.
Many workers will be going through the apprenticeship and the on-the-job programs
as the next few years go by. This is because the demand for concrete workers is
great because of the fast growth of the construction industry. If construction
activity continues at its rapid pace, all types of trained building tradesman will
be in demand.
As a cement mason, one could expect to earn $6.02 in11970. The range of hourly
wages ran from $7.80 down to $4.35 depending upon the city. In the rural areas a
cement worker's wage also varies much from one locality to another.
CEMENT WORKER
The mathematics involved with cement workers varies with the particular phase of the
job. The workers that go out to the job first and construct the forms for the
job must have a good knowledge of measurements and geometry. First of all, these
people must be sure that all work to be done is level or has the correct slope for
proper drainage. If a design of any type is desired, they must be able to construct
the forms for the desired effect.
For example, if a patio with redwood spokes in the form of a hexagon is desired, the
men must be able to determine the lengths of each side as well as the angle formed.
. Find m 44 A
2. Find length of XY
Find the area of 4 OXY
Hint: See the unit on "Polygons and Their
Areas" or your teacher to get help in
finding the area oft1OXY.
In this case, the men working on the forms also put in the order for the amount
of concrete needed. If the patio is to be 4" deep, find the number of cubic feet
(to the nearest 1') needed for this job.
Y
2
4. Total area of patio (total area = area of oneilx number of 41 's)
5. Total volume in cubic feet (V = total area x depth -- change units to feet)
(Ask your teachers or read the unit "Volume" for help if you do not under-
CEMENT WORKER
6. Total volume in cubic yards
When the forms are compl,ted, the concrete is ordered and poured. Finishers
smooth the surface for a mirror-like finish.
Another common job for a cement contractor, especially the smaller ones, is to
install front or rear steps. The small contractor will cut boards for forms
to the specifications in the drawing, and then figure the amount of concrete
necessary. Usually it is not a completely solid block of concrete; some fill is left
on the inside. If the contractor figures that the steps shown in figure 2 will be one-
third sand fill, how much concrete must he order for the following situation?
14
or,
14.1 t 8
14 H 81
_IL
A4,B C
D
1-< 72"
Sideview 32"
48"
Hint: Find the total volume of thL steps and subtract the fill.
a. Find total area of the side view (total of areas A, B, C, and D).
Total area =
b. Change area from square inches to square feet. A = sq. ft.
c. Find volume by: V = total area x depth (48"). V = cubic feet.
d. Change inches to feet and multiply. Now subtract one third of the total1
from your answer. This will give you the amount of concrete that must be
ordered in cubic feet. This must be changed to cubic yards. If 27 cubic
feet equals one cubic yard, how many cubic yards must be ordered?
Frontview
CEMENT WORKER
The G & P cement contracting company was fortunate enough to submit the low bid
au. the sidewalks aroulld a proposed aew park. The diagram of the project is
figure 3.
4I Fig 3
, CONCRETE WALK
8. If the sidewalks are to be 4" thick, how many cubic yards of concrete need
to be ordered?
1
1
CEMENT WORKER
G & P also was contracted fnr a triangular area in front of a library. Two different
Iideas were discussed. One allowed for a 6 x 8 x 10 triangle with a large
grass border. The other was for a similar larger triangle whose sides are three
times the length of the first triangle. As your next project, make a scale draw-
inging of each triangle. (Use4
" = l'scale on the graph paper provided.) Find the
cost differential of the concrete work between the two designs (assume G & P has
figured a total of $35 per cubic yard for material and labor),
Refer to these two triangles as you work problems 9 through 13.
9. What is the minimum relationship for two triangles to be similar?
10. What is the ratio between the sides of the two triangles?
11. What is the ratio of the areas of the two triangles?
Hint: Use Hero's formula for area
A = s(sii-b)(s-c), where s = 1 (a+b+c) and a, b, c are the lengths2
of the sides of the triangle
12. From 11, how many times greater is the cost of the larger triangle than the
smaller triangle?
13. What rule can be deduced about areas of similar triangles?
Those that work with cement can be put into two general classifications: the special-
ist -- one who works with a single phase of the operation, and the generalist -- one
who works with the project from start to finish. Geometry is an integral part of
both types, because each works with figures. The homeowner who wishes to save money
OE a home or patio or steps would hire a generalist who would need to use practical
geometry.
FORESTRY
For,'sters are concerned v'ith managing, protecting, and developing the nation'
forests and wildlife areas. ft it work covers a wide variety of activities,
such as mapping the location of timber estimating the amount of timber, super-
vising the harvesting and cutting of trees, and purchasing and selling timber.
They may also be involved in safeguarding forests from fires and disease, wild-
life protection, or managing camps, parks, or land.
You can see that much of a forester's work is done outdoors. The work is not
always easy and requires that n person be in good physical condition. Likewise
this work is often performed in remote. areas -- a forester must not object to
that. In addition, a "w,uld-be" Forester should have a strong interest in
science, for mnnv of the courses that he will be required to take involve
scientific concepts.
Thcse courses, as well is others, will he taken during college preparation
whi.(.1h is required of all foresters. A major in forestry is required and list
of colleges offering this kind of preparation is available in the guidance office
of your school. (enerally, courses in forest resources are also required r.,r
the forestry major.
As is the case in most careers today, specialization is affecting the forester,
too. More and more of the work now done H., the forester will he done by
specialist. The Occupational nutlook Handbook says, "Speciali7ed scient'-r-
biologists, horticulturists, agronomists, chemi,4rs, etc., increasingly
be hired for the more scientific worl: previously performed by forester.,
This means, of course, that the need for foresters may not increase as 7;uch
FORESTRY
as it would without these specialists. Government agencies (by far the largest
employer of foresters) will continue to need them as they expand their services
and as more land is put into recreational use.
Do you have an interest in nature and a desire to preserve it? Do you enjoy
working outdoors in a job requiring physical exertion? Do you have an interest
in science and in school so that you see yourself going to four years of college?
If you answer "yes" to these questions, it may be that forestry is a possible
career for you. Don't rule out forestry if you don't see yourself in college
either -- there are forestry aides and technicians that don't have to have
college training.
A forester's work involves many ideas and methods of indirect measurement
as will be seen in some of the problems that follow. An instrument commonly
used for field work in indirect measurement is the plane table, which consists
of a table mounted on a tripod. By means of a small "level," the table can
be set in a horizontal position. A small ruler with "sights" is arranged so
that the "line of sight" is parallel to the edge of the ruler. An object is
in the line of sight when it can be seen through the narrow vertical openings
in the sights. The diagrams on page three show how a plane table is used to
indirectly measure a distance, BC, which could not be measured directly because
a pond was between the points B and C. (See figure 1.)
2
FORESTRY
Figure 1
1. Suppose we set up the table at a point from which B and C can be seen
easily. After fastening a sheet of paper to the table top, put a pin in
the board at A. A ruler is placed with one edge against the pin and re-
volved to point in the direction of B. A line AD is drawn along the ruler
giving the direction of point B from A. The ruler is then revolved
about point A again, to sight in the direction of C. Another line, AE,
is drawn along the ruler, giving the direction of point C from A. Now
the distances from A to B and from A to C are measured directly with a
tape rule, For example, let the distances AB and AC be 400 feet and 600
feet respectively. (See figure 2.)
3
FORESTRY
Figure 2
Choose a suitable scale, say 1" = 25' and locate the points B' and C' on the4
plane-table drawing by marking off AB' and AC' to the scale on lines AD and AE
respectively. (Mark AB' and AC' off on figure 3.)
4
FORESTRY
Draw the line B'C' as shown on the sketch of the plane table in figure 1.
Since the length of B'C' in the plane-table'drawing is the scale length which
represunts the actual distance from B to C, the length of BC can be determined.
What is the length of BC?
The following picture shows a forest ranger using a plane table to determine
the location of a fire.
5
FORESTRY
2. Two forest lookout stations, six miles apart, are at points A and B as
shown below. Forest service personnel in each tower observe a fire located
at point C. The observer at lookout A finds that the measure of the angle
from the fire to lookout B is 77 degrees. The observer at B measures an
angle of 53 degrees from the fire to lookout A. Row far is the fire from
lookout A to the nearest tenth of a mile? a.
Hint: one way to solve the problem is to construct a scale drawing and
measure the distance. Use your ruler and protractor and construct a scale
drawing.
How far is the fire from lookout B? b.
Let your scale be 1/16" = .1 mile
6
FORESTRY
3. Two camps were located at points A and B on opposite sides of a lake as
shown below. Some measurements are indicated on the figure. (m <BAD = Tu.-CADE)
-/---\/ .... IIgam' ...... I
A III0°.0.0.0
0.00
10 yd
B
a, Whit additional measurement must be made in order to determine the
distance, AB, across the lakr,7
b. Make a scale drawing and find the distance AB to the nearest yard.
AB.= Let 1/16" = 1 yard.
FORESTRY
There are times when a forester may have to use indirect measurement, to find
the height of a tree for example. The next problem would be one method he
could try.
4. How high is the tree if it casts a shadow of 38' when a man 6' tall casts
a shadow of 11'? Height =
(Use proportions from similar triangles.)
---04381t.
5, A ranger measured the circumference of a tree at waist height. If the
1circumference is 169pn" what is the diameter of the tree?
Diameter =
8
FORESTRY
6. The distance between two points on a cliff is to be measured. The
rectangle method" is used as shown in the figure. r
a. What kind of angles are formed at A and B?
b. What is true about the distances AD and BC?
c. What distance should be measured to determine indirectly the distance
AB?
The compass has aided foresters in finding their way in the woods for many
years. When in proper operation, the needle turns until it points to the mag-
netic North Pole. A forester would revolve the compass until the zero on its
scale is placed directly at the "north" end of the needle as shown in the dia-
gram.
Now by sighting across the center of the in-
strument to some distant object and noting
where the line of sight crosses the scale,
the forester can read the magnetic "bearing,"
or direction angle, of the object from the
position he is in. The observed bearing of
A is 20 .
9
FORESTRY
7. What is the observed bearing of
a. B?
b. C?
c. D?
Notice that with this kind of a compass, all bearings are read "clockwise"
from the north.
For further information write:
American Forestry Association919 Seventeenth Street N.W.Washington, DC 20006
Society of American Foresters1010 Sixteenth Street N.W.Washington, DC 20036
10
ELECTRICIAN
Jake's folks decided to build an addition on to the house. They had always
thought that a recreation room and extra bedroom would be nice. Jake and his
dad had done some of the work, but now it was time to put in the wiring and
Jake's dad didn't want to tackle that himself. An electrician was hired and Jake
got to know him quite well while he was watching hime wire the new addition.
Much of the time they talked about the electrician's work. Jake discovered
that the electrician did most of the work the way he did because of
codes. These are rules that the state and local governments have adopted
to insure safe, attractive, and unifotm electrical installations. The out-
lets that were being installed, for instance, had to be properly grounded.
The electrician told Jake that he was self-employed but that most of the elec-
tricians he knew worked for a contractor. He was therefore a little more "on
his own," but at times had more trouble finding work because he didn't have the
large construction jobs the bigger contractor would get.
Jake asked his dad how much the electrician was going to charge. Jake's dad
said that would depend on how long the job took and that all of this talking to
the electrician didn't help any. The company was charging $18 an hour plus all
the supplies, of course. Jake figured that a couple of days' work by the electrician
could really add up. He discovered, however, that $18 an hour was about the average
of what was charged in only that part of the country.
Jake had two more questions for the electrician. The first concerned the future
of this line of work and the second dealt with the theory of electricity. Jake
was told that the electrician of the future could be doing quite a different job.
Already a couple of the electrician's friends had gone to work for a firm that made
ELECTRICIAN
modular or packaged homes. This meant that the work could be done in a "factory
style" setting with the wiring and electrical work installed in almost an assembly
line fashion. The electrician said this meant faster work under better, mor:-
uniform conditions but would be even less appealing to him than working for a large
contractor. Jake could understand that this electrician wanted to be more on his
own, with the freedom that being your own boEs provides.
The electrician then explained some of the theory of electricity to Jake. "[deter
is measured in gallons, meat in pounds. Electrical power cannot be weighed on a
scale because it is in constant motion. Instead, electricity is measured by how much
flows past a point in the electric line at any given moment.
"When we want to measure a quantity of water, we talk of 'gallons.' When we
talk of water in motion, we talk about 'gallons per second.' With respect to
electricity, a unit of electricity in motion is called an ampere_.
'When air is put under pressure, we speak of pounds of pressure per square inch.
Electric power is also under pressure. This pressure is measured in volts. A
small radio battery has two volts of pressure, regular house current has 115
volts of electrical pressure. DD you know how many volts a car battery has?
Amperes or volts alone do not determine the amount of power flowing in a wire,
the two combined do. Watts is the term used to determine the amount of power
(amperes and volts) running through a line. 'Watts = Volts x Amperes.'"
The electric:.an used the example of a rifle. If the speed of a bullet remains
constant (volts) but the weight of the bullet varies (amperes), the impact
varies. The opposite is also true. If the weight stays constant (amperes)
and the speed changes (volts), the impact varies.
2
ELECTRICIAN
Much of the mathematics an electrician uses is applied higher algebra and tri-
onometry. He must be able to read blueprints, much of which are in the form
of geometric figures. Applied geometry, especially working with angles and
missing parts of geometric figures, is a must.
An example of the applied use of the Pythagorean Theorem is shown in the follow-
ing problem.
1. A 60' length of conduit is bent as shown. Find the distance of X.
20 isE
22'- - --3
X = OFFSET
I
Noticed BEC is a rights. The length BE is known, BC can be found from the
diagram (AB +BC+CD= 60),and from this X (same as EC) can be calculated.
Pythagorean Theorem
Find X:
A
3
; C2
= a2
-V b2
Where C is the
hypoteneuse (side opposite the 90° angle)
and a and b are the other sideD
ELECTRICIAN
Another application of the Pythagorean Theorem is in finding the voltage
across two vectors.
2. Given: a 100. -volt line between
L. and L3 2
a 144 volt line between
L and L2 1
Find: the voltage between L and L3 1
/L1
144 voltsb
L3
LTL2
100 voltsa
1
Notice that the line between L. 3 and L. 1 would be the hypoteneuse of a right
triangle. Thus you can use the formula C2 = a2 + b 2 .
Find C (approximate).
4
ELECTRICIAN
An electrician must also work with circle measurements. The following
diagram shows an applied situation. The diagram is of the inside of a
roof meeting an inside ceiling. He must figure what length of pipe to
use, including the bend, to go from A to B. The electrician has found the
center of the circle whose arc is DC and radius is 3-1-2-".
Find the total circumference of the circle 0, then find the length of arc
DC. Let ir = 3.14.
3. :.) Circumference of circle 0 = fr77 D =
h. L360-
ength of arc DC = -115,°, of C =
c. Total length of conduit needed to go from A to B =
5
ELECTRICIAN
The electrician must also be a good carpenter. His knowledge of basic
geometry is essential to his everyday work.
I
6
GENERAL CONTRACTOR
Sam ,:anted to work in many construction fields, lie didn't want to limit
himself to just one, such as carpentry, plumbing, or electrenies. Yet he
knew that it would be difficult to become skilled in all the trades, not to
mention the tremendous cost of being equipped in so many occupations.
So he decided that maybe a better solution was to become a general contractor.
A general contractor takes responsibility for the overall construction of
building. Much of the work lie is directly responsible for. Some phases of the
work, such as electrical work or plumbing, he contracts other tradesnwm to do.
While looking through the math he should be responsible for knowing, he found
that he must know how to make accurate'e,stimates of a job. He must have enough
knowledge of other trades to be able to give good bids for a ioh. He then
contracts other tradesmen to do various phases of construction that need to be
done and which he is not equipped to do. Here are some examples of problems
general contractor miglit have to do.
l. The basement of a house 28' long and 24' wide is to be excavated to a depth
of 6'. A building ordinance requires that the excavation be extended 11
beyond the wall on each side to permit inspection of the masonry. At
per cubic yards, find the cost contracting an excavator. Cost
(Volume Lx Wx1)
2. The house in problem one is located on a lot 170' long and 45' wide. The level
of the lot is to be raised 18". Find the cost of filling at $.90 per cubic yards.
Cost =
GENERAL CONTRACTOR
3. Both sides of the roof shown are to be covered with asphalt shingles.
Find the number of sq. ft. of surface to be covered.
42'
N
Number of sq. ft. = a.
If 100 sq. ft. = 1 sq. of shingles and if 1 sq. of shingles comes in 3
bundles, then how many bundles must be ordered? b.
Adding one-third square for a Boston Ridge, how many bundles must be ordered in
all to complete the job? c.
A contractor can sub-contract a shingler to put on the roofing for $5
a square. What will the total cost be? d.
4. A contractor must estimate the excavation and rock fill needed for two
circular dry wells. How many cubic yards of rock fill will it take if one
well has a diameter of 5'-0" and a depth of 6'-6" and the other has a dia-
meter of 5'-6" and a depth of 7'-0"? (Vol. of cylinder =17r2 x H)
Cu. yd. =
5. For estimating purposes, when computing labor costs for finish flooring,
the actual area of the floor, in sq. ft., used. An average carpenter
can lay, scrape the joints and sand approximately two squares (200 sq. ft.)
of finish flooring in an eight-hour day.
GENERAL CONTRACTOR
1 long 1, i 11 it take to lay finish flooring in the room shown below?
Hrs. = At $6.80 per hour, what is the labor cost?
Cost
12'0
1
I
1 I
1 !
1 1
1
14 0-
Read the blueprint on the next page.
n, What is the length of the house measured along the west wall'
Length =
b, What is the perimeter of the house:
Perimeter =
c. How many joists will be required for the south span?
Number of joists =
d. At $6.04 each, how much will the joists for the south span cost?
Cost =
3
GENERAL CONTRACTOR
II
0
0(8% LT.P GIRDER
2 ...17.:11rV1171-270:31.
6"X e" L.L.T.R GIRDER= =
2=4"
:160
24
9)1.0 i-=
x
LINE OF ROOF RIDGE
611 - 7 3 .4-- 6:4"
0
I
Sometimes it is necessary for you to find the urea of a triangle tie
knowing the height. If you know the lengths of the three sides are a,
and c, then use the following formula:
Area of triangle =1(s(s-n)(u-b)(s-c T
where s = 1., (a+b+c)
4
GENERAL CONTRACTOR
7. Some triangular-shaped land was offered for sale at $3,000 per acre. A
contractor was interested in buying the land to plot into smaller lots
and build homes on them to sell. if the sides of thr land were 50 rods,
90 rods, a"d 100 rods, find the area of the land in sq. rods.
Area =
Since 160 sq. rods = 1 acre, what is the sale price of the land?
Price =
8. Concrete is obtained by mixing 4 parts gravel, 3 parts sand, and 1 part
cement, by volume. Determine the number of cubic yards of each material
required to make 150 cubic yards of concrete.
Gravel =
Sand =
Cement =
A porch roof is to be supported by two solid concrete pillars. The pillars
are each 2'-0" in diameter and 10'-0" high. How many cubic feet of concrete
will the two pillars contain? (V -41-'x r2 x H)
At $21 per yard (cu. yd.), how much will this cost?
Cost =
5
GENERAL CONTRACTOR
GJ;:1-)te ti w_.fact :u.:2a to ;!c floored in the s(!mi-circular hay window
11 flow many sq. ft. of sdHlooring ;Ire recinird L, cove': all the floors ot
the rooms below':
Area
2 01 0"
6
HOME PLANNING
"Home's not merely four square walls -- home is where affection calls." "What's
that got to do with all tnis?" Tom asked Judy. They were trying to design the
living arrangements for their new home and Tom was determined to slap up four
walls and be done with it. It would be much cheaper and they could make use
of their present furniture, rather than buy new units for everything. He
was exasperated with Judy's constantly changing ideas on arrangement. Tom
preferred having the fixed items built-in since that would give a neater arrange-
ment cis well as have the main part of the floor in any room free for any group
type nrtivities they might wish to plan.
The first thing they had done was contact an architect for the basic layout and
plans. Then they could decide whether they'd do the work themselves or contract
it out to one of the many registered remodelers available.
First Tom and Judy tackled the kitchen. Many basic arrangements were available
depending upon what their home and entertainment ideas were, from the one-wall
arrangement common in apartment buildings and small homes to the most popular
U-shaped kitchen. In all cases, though, they had to be conscious of the work
triangle connecting the range, refrigerator, and sink centers. For maximum efficiency,
the triangle perimeter should not be more than 22 feet. Doors should he located
so that traffic lanes do not cross the work triangle. It is generally accepted
that distances between appliances should be four to seven feet between refrig-
erator and sink, four to six feet between sink and range, and four to nine feet
between range and refrigerator.
HOME PLANNING
Below you will see some of the arrangements available to Tom and Judy. In
each case, you are to:
A. Draw the step-saving triangle.(Use the points given for accuracy.)
B. Label the sides a,b,c.
C. Measure a, b, c (scale IP = 1')8
D. Compute the perimeter:. P = a+b+c
E. Compute the area by using Heron's formula: A =-1s(s-.a)(s-b)(s-c)
where s = 1 (a+b+c)
I
S REF
1.RANGE
1. U-shaped kitchen
A. Draw
B. Label sides
C. a =
D. P =
E. S =
FR7-
ovEN
b = C =
; A =
II
III
2. Island Arrangement
A. Draw
B. Label sides
REF
OVEN
fs]
Esj
3. Peninsula Arrangement
A. Draw
B. Label sides
C. a = b = C = C. a= b= c:-...
D. P = D. P =
E. S = ;A = E, S = ; A =
2
IV
re
L_
HOME PLANNING
V VI
OVEN
REF
4. L-Jhaped 5. Peninsular 6. Modifier? U-shaped
A. P = A. P = A. P =
B. A = B. A = B. A =
Tom and Judy felt the organization of the kitchen was most important
since it would be used in most activities in the housL for both family
living and entertaining. They could not afford a second kitchen in the
recreation room, so their main kitchen would serve double duty. Really,
this was Judy's opinion and Tom thought best to agree with her, especially
since he considered her better skilled at home planning than he.
The layout of the entire home was tackled room by room and many different dec-
isions-had to be nude. Furniture arrangement had a greater importance than Tom
first realized. Until the architect explained flow patterns to him, he had
considered Judy's occasional furniture re-arrangement sprees merely exas-
perating feminine whims.
3
HOME PLANNING
To illustrate without getting into lengthy discussions, consider the sleeping
arrangements for Tom and Judy's two girls. They had to share a bedroom so one
room would have to contain two single beds as well as a small chest of drawen;
and a vanity for each girl. Needless to say, Tom and Judy would also like to
keep as much available floor space for the girls to use. A glance at the two
arrangements shown should clarify the points being made.
The usual side-by-side arrangement of twin bedsleaves little open floor space. In many rooms,
the beds can be separated to leave more openfloor space, and each person has his own area.
Another problem that Tom and Judy had was that their house was located on a rather
busy street. It would be desirable not to have to back out onto the street where
there would always be a possibility of accidents. Two basic plans seemed to suit
their purposes. The first was a circular drive which might be added to the straight
drive heading to the garage for a turnaround; the second would be a spur added
to a straight driveway; this would provide additional off-street parking or space
for turning around in order to head into street traffic.
On page five are sketches of the two possibilities and you are to constrict accurate
layouts of the circular drive and the drive with the spur addition. The measurements
4
HOME PLANNING
are all given below and you are to use a scale of 1/8" = 6' for your construction.
In some cases, six feet does not divide evenly into the dimensions needed, so
you will need to first calculate the fractional part of an inch needed.
HOUSE A GARAGE HOUSE ii GARAGE
S IDE WALK
21.-11
0,O'
0
0,
m
SIDEWALK
ROAD WAY ROAD WAY
Specifications
Driveway: 18' wide =
ScaleDimensions
_F
ScaleDimensions
(Answer in inches) Specifications (Answer in inches)
3Example 7.
Length of main drive: 78' =
Circular drive:
Inside radius: 18' =
Width of drive: 12' = 10.
Outside radius: 18' I. 12' = 11.
Driveway: 18' wide = 14.
Length of main drive: 78' = 15.
Spur:
9. Radius of curved portion: 9' = 16.
Length of non-curved portion:9" =
Width of non-curved portion:The circles are concentric and their 12' = 18.
center lies on the ;louse side of thesidewalk; also, the inside circle *Curved portions of spur should beshould be tangent to the main driveway. tangent to both main drive and non-
curved portions of the spur.
17.
Exit of circular drive:
Width: 12' =
Length: 15' =
General appearance:
12.
13.
LI r
3 C._
5
General appearance:
HOME PLANNING
Tom and Judy wondered about the relative costs of paving both driveways. In order
to determine that, they would have to know the total area of the drives and then.
assuming a thickness of 4" of concrete, find the volume of concrete needed in
cubic yards. In order to see what calulations would be involved, determine th,.
surface area of the circular arrangement only. For our purposes, don't worry
about the part of the circular drive that overlaps with the main drive. The
extra concrete ordered because of this will be used for fill in back of the house.
Ilse the following outline to aid your thinking:
19. Length of main drive:
20. Width of main drive:
21. Area of main drive:
22. Radius of larger circle:
23. Area of larger semicircle:
24. Radius of smaller circle:
25. Area of smaller semicircle:
26. Area of circular drive:
27. Total area of drive:
28. Volume of drive (in cubic feet):
29. VoThme of drive (in cubic yards):
30. At $21 per cubic yard, cost of paving drive:
HONE PLANNING
Tom and Judy also calculated the volume of the drive with the spur to compare
the cost. By calculations similar to the one outlined above, they learned that
driveway B would require 201 cubic yards. At $21 per cubic yard, that layout
would cost 31. . They realized that money would not be the only
consideration in choosing which drive to install. Which one would you choose,
both for convenience as well as cost? 32.
As is easily seen, all these improvements would be costly, but Tom and Judy
thoroughly enjoyed planning the home. They realized that working with an
architect and a trained home economist was crucial for the end result to be
both efficient and enjoyable. Only part of the planning which goes into a home
could be illustrated here, but it is hoped that enough was given sc that you
may, also have some ideas when, in later years, you plan your home.
7
CABINETMAKING
After Bill finished exploring the use of geometry as a carpenter, he got to
thinking that the math involved wasn't all that bad. The thing that bothered
him was working under outdoor conditions. When the weather is nice it's great,
but you can't always depend on nice weather. When you consider winter, rainy,
and windy days, there might be some miserable days to work in and possibly even
weather lay-offs.
Bill did like woodworking, though, so looked into cabinetmaking for a career.
the work requires creativity and precision with hand tools. Besides, it would
be inside work.
Bill found much similarity between the jobs of a carpenter and a cabinetmaker.
in fact, many cabinetmakers start out as carpenters and many others work in both
fields, depending upon the needs at the moment. The work seemed to require a
little more precision and almost the whole story of a finished product seemed
to be found in the way the cabinet looked when finished. Bill liked that, for
making "something just perfect" was fun. He found the pay to be much the same
for both trades and most of the other factors he looked into overlapped between
the two.
A big question for Bill was, "Do I have what it takes to be a cabinetmaker?"
For help in answering this question, he went to see his counselor. The counselor
looked at Bill's grades and found that he had done quite well in shop courses.
In addition, some tests showed that Bill had much better than average aptitude
in mechanical and spatial areas. The counselor asked Bill about his hobbies
and found that he liked to do creative things, especially those which involved
CABINETMAKING
working with his hands. It seemed to the counselor that when all of these thinc-;
were put together, Bill had the ability and the interest to do the work require'
of a cabinetmaker.
A second question was, "Would I have more difficulty with mathematics as a
cabinetmaker? Bill decided to find out what types of problems he would encounter
while making cabinets or furniture. Here are some examples of math involved.
1. Suppose you/see a picture or a sketch of a furniture piece you would
like to make. In many cases the three major dimensions of height, depth,
and width are given. With careful planning you can design a product that
is very similar. Consider the following diagram on the next page and
follow the directions using figure 3.
a. Fasten the picture or sketch to the left edge of a piece of paper.
b. Use a T-square to extend the straight lines from the edges of the
furniture piece.
c. Suppose that the overall height is 30". Select a scale in which 30
units equal in length will fit diagonally between the two horizonal
lines. Place the first division point on the scale on the top line
and the last on the bottom line.
Now extend the lines (in b)until they intersect the diagonal line for each
dimension needed. Now you can count the number of units or inches
for this part. The same method can be used to find other dimensions.
e. Find the following dimensions: (See diagram on following page.)
A=B=C =D=
E =F =
G =H=
2
CABINETMAKING
FIG. I FIG. 2
A
FIG. 3Determining the dimension of the ports of this chest. The overall size is 30"W x 15"D x 30"H.
30
2. A bookcase is arranged so that the spaces between its shelves are 10",
16", and 8" respectively. If a slanting board is nailed diagonally
across the back for re-enforcement and cuts off a segment of 25" between
the first two shelves, what segments are cut off between the remaining
shelves? (Hint: Use ratios; see "Parallel Lines" unit.) \ .13
a and b.\.
r"
3
CABINETMAKING
3. An octagon is often used for wood products such as wastepaper baskets and
small tables. What would be the angle of the cut for the edge of each
face of an octagon end table if you make a mitered cut? (Assume the oct-
agon is regular.) Hint: find the measure m° (See unit en polygons.)
and take 1 of measure m°.2
1 m°7
4. Many irregular shaped objects such as Early American furniture pieces
have arcs, circles, and lines which are tangent. Arcs or circles are tan-
gent when they touch at only one point but do not intersect.
a. Given the radii of 3 arcs as 2 in., 1 in., and 1/2 in., construct a
series of three tangent arcs. Use given point 0 as the center of
the arc with a 1 in radius so that the three arcs are tangent in a2
series.
P
Make all the construction marks light and show the arcs in bold lines.
4
CABINETMAKING
b. Draw an arc with a radius of 1 inch tangent to the given line AB
and the given arc CD. The center of arc CI) is K.
A 1.3
5. What length of 1" veneer strip is needed to go around a round table with
a diameter of 54 inches?
Usolr = 3.14.
6. If you used a 2'-()" square piece of plywood to make a 2'-0" diameter
round table top, how much wood is wasted?
Sq. ft. wasted = (See unit on polygons and areas.)
7. Projects found in books and magazines are seldom drawn to full size. If
the product contains irregular parts, it is necessary first to enlarge
these to full size or scale. In the scale drawing,on the following page,
of a corner shelf, consider each square equivalent to a 1" square. Make 1" squares
and transfer the design of one shelf or one side to the 1" squares. (Hint:
continue to locate and transfer points until enough are marked for a pat-
tern. Sketch the curves lightly freehand, then use a French curve to
darken the lines and to produce a smooth, evenly curved line.)
CABINETMAKING
r4ANGER
8. On a drawing of a scale of 1/16" = 1'-0", what
meat of a line representing 1001-0" ?
9. A circular table top is broken and the loose piece has been lost in
would be the scaled measure-
shipment. You have been asked to duplicate this top. You have an idea
where the center of the diameter is. Show on the diagram (on the follow-
ing page) how you would locate the center to duplicate the table top.
(Hint: work with the perpendicular bisectors of a chord.)
6
PLUMBING AND PIPEFITTIN(
Many people link plumbers and pipefitters together; there are some like-
nesses, but there are many differences. A plumber installs and maintains
piping systems which carry water for domestic and commercial uses. Installa-
tion cf waste pipes and plumbing fixtures as well as gas lines are his main
concerns. A plumber must be able to select pipes, valves, and other fittings
appropriate for the job and state and local codes.
A pipefitter-steamfitter works in such new areas as atomic energy power
houses and total energy systems as well as the older steam systems of heat and
enexgy. He must apply a technical knowledge of 'welding, brazing, soldering,
bending, and threading in the installation and maintenance of piping systems.
He must have a working knowledge f the various types of power plant installa-
tion as well as heating systems and process piping systems.
An individual interested in becoming a plumber or pipefitter can go about it
in a couple of ways. There are at least four vocational-technical schools in
Minnesota where an individual can get training. The course is 10 months long
and is available at St. Paul, Jackson, St. Cloud, and Wadena. This schooling
will cut down on the length of time that a person must serve as an apprentice.
0,Thers may start out as apprentices without trade school preparation. They
begin work at 40 to 50 percent of the journeyman's wage and need to spend five
years in the training program. During this time, they will take realted class-
room training. There are 350,000 plumbers and pipefitters at the present time
and this total is expected to increase as constuction activity continues to
increase. These tradesmen earn an average of $7 per hour and can earn as
PLUMBING AND PIPEFITTING
much as $9.42 as plumbers in Oakland, California, do. In addition, they usually
receive many desirable fringe benefits.
Ability in areas of physics and math are more important than one would think
because many applications of the angle, volume, and circle principles are in-
volved.
People working on both plumbing and pipefitting must have an understanding
of the carrying capacity of differcnt size pipes.
Example: A 4" water pipe can carry how many times as much water as a 1" pipe?
Solution: A quick answer would be four times, but let's take a look at a
drawing of the two.
As you can see in the dia-
gram, many more than four 1"
pipes are contained in a 4"
pipe. To see the solution
geometrically, we can com-
pare the areas of both pipes.
End View of Pipes
2
PLUMBING AND PIPEFITTING
22A = x radius squared, if = 2 . ih or 7
area of 4" pipe 4.22 /,/ _22 4 , 16
area of 1" pipe /r2 12
1
2 4
Therefore the 4" pipe carries 16 tim.-:s more water than the 1" pipe.
1. Find how many times as much water a 6" pipe carries than a 1" pipe.
1 ,
2. How much more water will a 6" pipe carry than a 12"
3. An 18" main steam pipe coming from a boiler can have a number of 4" branch
pipes from it to equal the area of the main. How many 4" branch pipes
can be installed?
Hint: Use same idea as in previous example. Determine the area of the
main pipe and then the area of one 4" pipe.
3
PLUMBING AND PIPEFITTING
4. Three separate pipes have been used to carry drainage from the second
1floor of a house. If their diameters are 2", 3", and 1 ,, , find the9
diameter of a single pipe having the same carrying capacity. Make a
drawing of the single pipe.
Diameter
By douOing the inside diameter of a pipe, how many times does the
capacity increase? Choose at least two examples of pipe to double.
6. The diameter of a piston n a steam pump is 8 inches. When the steam
pressure is 80 pounds per square inch(PSI), the total pressure on the piston
is? (Determihe the number:' of square inches of the piston then multiply by 80 PSI.)
Pressure =
4
PLUMBING AND PIPEFITTING
Whenever the run of a pipe must change directions, the angle formed by this
deviation is known as the offset angle.
Example
R
27!
S!OFFSETANGLE
5
PLUMBING AND PIPEFITTING
Notice that if we use S, R, and H from the diagram, a right triangle is
formed.
Common offset angles are 300, 45°, and 60°. These situations, then, are
examples of the practical applications of a 30°-60°-90° and
45o_45o...90o triangles.
Recall: the ratio of the sides of a 300-600-90° is
and of a 45°- 45° -90° is
7. A plumber must adjust his run so that S = 10" and the offset angle is 30°.
How long a piece of pipe (length x) must be cut for the distance H? (Refer
to the 30°-60°-90° triangle.)
Find the offset angle A if the height is 5" and the run is 5". (See diagram.)
(Refer to the 45 °- 45 ° -90° triangle.) m4:A a.
What length pipe is needed for H? b.
6
PLUMBING AND PIPEFITTIN7
A more difficult offset is the perpendicular offset. It is one in which the
run after the offset is perpendicular to the rur, before the offset.
Notice that pipe A is "perpendicular" to
B (actually, their planes are J.). If the
pipefitter knows that the distance C
is 36", how long a section of pipe is
needed for the diagonal run?
8
7
back bottom corner
NOTE: dotted lines are bottom lines
PLUMBING AND PIPEFITTING
For the job pictured at right, the pipe-
fitter must find the diagonal of a rec-
tangular solid to find the length of the diagonal run.
If A = 6'
D = 8'
B = 10'
Fig. 1
Step 1
Find the length of the diagonal of the bottom
rectangle. Use the Pythogoran Theorem E2 = A2+D2
9. E=
Step 2
Use E from fig. 2 as the side of a triangle and find C.
10. C = (length of diagonal run)
Fig. 3
45'
The plumber and pipefitter must spend hours of preparation in math. But the
90'
8=40'
learning process is much easier because they are working with material and realize
that this will be their occupation. After a few year on the job, they develop
"a feel" for their work and can readily cut pipes and choose fittings for the
proper installatIton.
8
SURVEYOR
One day during lunch, Chuck was talking to his friend about summer jobs. Chuck
had been wondering what lie was going to do. He hadn't worked previous summer,.,
but now lie needed some extra money for his senior year.
His buddy told Chuck to check into getting a job as a rodman with a surveying
crew. That was the job he had, and thc, money wasn't too bad. Chuck asked his
friend to tell more, especially to get a iob like that.
He found that a rodman holds the pcie, assisting in measuring the height
or distance or direction that a surveyor is working. Chuck was told that
most jobs like this are with a government agency (county, city, or state)
and the surveying crew is "laying out" roads, etc.
Chuck was concerned -- he didn't know much about surveying and would have to
learn everything; his friend said that he had been in the same situation and
had been taught on the job. This summer ChucF's buddy was going to be a
chainman -- measuring distances, instead of holding the pole.
Chuck talked further and found that many people become surveyors after starting
as rod or chainmen. Others go to a trade school for one-and-one-half years
and get training in something like highway engineering. Many surveyors go to
work for a government agency at a salary of either $7,300 or $8,100 a year with
increases determined by experience and examination results. Private firms
negotiate wages depending on need, experience, schooling, and other factors,
but are usually close to those paid by the government.
SURVEYOR
Chuck found that when a surveyor makes a "plot" drawing of a piece of ground
he has surveved,he has to make a great many calculations. He must take bear-
ings (measurements in degrees) from various extablished points to lay out the
plot. A north-south line is used as the standard and all readings are taken
from it.
Study the following four figures.
AN N
11 E 1V
51'
S S
E
C
61'
D
S S
In figure A,the bearing of the line indicated by the arrow is 65° east of north,
figure B,51° west of north; figure C 52° east of south, and figure D, 610 west
of south. For convenience, these bearings may be written fig! A, N 65°E; fig.
13, N 51°W: fig. C, S 52°E: fig. D,S 61°W.
Read the bearings of the lines marked by arrows in E through H.
E
N
1. Fig. E
2. Fig. F
F G
N
3. Fig. G
4. Fig. H
H
SURVEYOR
Surveyors must be able to read blueprint drawings and descriptions of pro-
perty lines. A blueprint is a scaled drawing showing the length and direction
of borders of property or the direction, widths, angles, etc. of roads. The
scale used on the blueprint that ia the unit of measure to be used on the blue-
print for each actual unit of measure of the project, is determined by the size
of the project and the desired size of the paper the blueprint is on. The
amount of detail needed is also a factor in the size of the scale.
Excmple: A rectangular plot 16' by 24' could be scaled down to 16" by 24"
on the drawing. The scale would be 1" = 1'. If the scale was 2"
= l', the drawing would be 32" by 48". If another plot of land
was 75' by 30' and the scale for it is 1" = 5', the scaled drawing
would be 5. a
Find the actual length of each side of each of the following
figures. Use a ruler to find the lengths of each line segment, then
determine the actual length by using the given scalefig. A fig. B
A
Scale 1" = 10'
b. Actual size of figure A L.- w =
c. Actual size of figure B AB= BC=
CD= DA=
3
Scale 1" = 12°
SURVEYOR
6. Combining a knowledge of bearings with that of scales, draw a diagram of
the following property. The description begins from point A (northwest
corner of property). (Use a protractor for measuring angles, a ruler for
urits. The scale is 1" = 50'.) From point A go N 450 E 100'; then S 900
F 150'; then S 45° E 125'; the S 00 100'; then N 85° W 300'.
a. From the last established point, find the distance to point A (approximate
if not exact).
4
SURVEYOR
e. Also, using a protractor, find the bearning from the last established
point to point A.
Below is a drawing of a piece of property. Give a description of the property,
as in the last problem, by giving directions and distance from each point start-
ing at A. Scale: I" = 50'
N
Li
5
i
7. Decription of above property:
SURVEYOR
8. What is the distance (approximated using the scale and your ruler) from
point Z to A?
9. Find the bearing reading from Z to point A.
Chuck felt the practical application of geometry in surveying was much more
interesting than studying geometry from a book. The on-the-job aspect seemed to make
geometry more useful.
6
OUTDOOR ADVERTISING
Bill came out of his geometry class shaking his head and muttering to himself.
He had just learned that his semester grade would be dependent on a project on
the uses of geometry in industry. Only last week EJ Zilensky, the class
agitator, had tried to put Mr. Meier on the spot by asking him what good was
anything they were studying. Mr. Meier had talked for some minutes on prac-
tical applications of geometry, but Bill didn't take notes because most of what
he said wasn't too important so he thought. Now he wished that he had taken
notes, for they would have come in handy. Mr. Meier had earlier told the class
to expect a project of some sort, but the students were keeping quiet about it
in hopes that he had forgotten. It must have been Ed's remarks that brought
up the whole matter again.
Practical applications of geometry. Like what? Bill thought more about it as
he drove home. He'd like something different and interesting, too! Casually
he began humming "It's the real thing "That 's it!" All around him
he found billboards advertising all sorts of products. "Why not research the
uses of geometry in outdoor advertising?"
Bill's first step was to contact one of the billboard companies in town. He
was able to tour the plant and learn first-hand how billboards are produced.
In addition, he learned that 5!.dvertising firms employ all kinds of different
workers from executives to artists and layout specialists, salesmen, copywriters,
and other administrative people.
1.
Bill found that advertising firms have many different people of very different
OUTDOOR ADVERTISING
backgrounds earning extremely different salaries. They all, however, seemed
to have artistic and language ability and interest in addition to an intense
interest in working with people. These seemed to be the things that made for
successful advertising people.
Managers, account executives, copywriters, media directors, production managers,
and research directors were some of the people that Bill saw at work. They
all make up a part of the 140,000 nv....n and women working in the area of advc,--
tising. This number will not increase significantly during the next few years
and competition for jobs will be keen through the 1970's. The highly qualified
person, however, will always be in a good position to find employment in this
field. "Highly qualified" means being experienced, creative, imaginative, and able
to get along well with co-workers. This kind of person, Bill decided, would
always be in a good position to find work.
As many as 120,000,000 people may view a poster in one day, depending on its
location. Exposure is obvious and exposure is extremely important if one is
going to sell a product. Secondly, communications is the name of the game.
People need to be convinced that this product is really worth buying, and this
communication must take place rapidly, since they're moving and will shortly
move out of sight range.
There were basic principles to be followed and, though these were not strictly
geometric, Bill thought he should list them:
2
OUTDOOR ADVERTISING
7 Principles of Outdoor DesignAt 'each step of development, from start tofinish, check your design against these sevenbasic principles:1. PRODUCT IDENTIFICATIONDoes the advertiser's name or productregister quickly?2. SHORT COPYIs the basic idea expressed quickly,and with impact?3. SHORT WORDSCari the reader grasp the idea at a glance?4. LEGIBLE TYPECan the reader read the copy at a distance,while moving?5. LARGE ILLUSTRATIONSAre the pictures big as all Outdoor?6. BOLD COLORSAre colors clearly defined? Do theyhave impact?7. SIMPLE BACKGROUNDDoes the background interfere with thebasic idea .. . or help it?
So far, Bill had been able to gather much interesting informar4on but little
which could be related to geometry. Outdoor advertising was definitely a form
of art and governed by the rules of art more than by other more rigid
guidelines. Color and contrast played very important roles in getting the
message across in addition to the copy used, and some of this would be directly
related to geometry.
However, the names posters and bulletins kept cropping up and Bill soon learned
that there was a difference between them, although various c;esigns could be
used for both with minor alterations. Posters are prepared in the proportion
11 x 2.-
4while bulletins are usually in the proportion 1 x 31 measuring
2
14' high and 48' long. As you can see, this proportion isn't
3
OUTDOOR ADVERTISING
exactly 1 x 11 for the bulletins but rather 1 x I. ; for practical
purposes, though, 1 x 32 close enough.2
Remembering that posters run in the proportion 1 x 23-- determine the cor-4
rect sizes of copies submitted to the advertiser for approval: these usually
run 6" x 2. or 8" x 3. Finished artwork may be prepared
in 12" x 4. It i6" x 5. ", or 20" x 6. ", all of which would
maintain the same proportion. In many instances, a design may appear on both
posters and bulletins.
Posters come in three basic varieties:
I. 24-sheet poster
Originally, when presses were smaller, 24 sheets of paper were required for
this. Larger sheets-re now used, but the term remains. The area between the
design and the frame is covered with white "blanking" paper.
4
OUTDOOR ADVERTISING
II. 30-sheet poster
This gives about 25 percent more area for a design. Notice that the width of the
"blanking" paper'is quite reduced.
III. Bleed poster
This is the same as the 30-sheet poster except that the design is carried all
the way to the frame. This t done by printing the "blanking" paper and offers
40 percent more design area than a 24-sheet.
5
OUTDOOR ADVERTISING
Outdoor posters are reproduced on paper by means of silk screen, letterpress, or
lithography, depending on many factors, including the number of posters required.
The three standard sizes have gpecifications given in the following table Use
the information given to figure the total exposed area for tht three types of
posters and then check it against the standard poster ratio.
24-SHEET 30-SHEET BLEED
COPY AREA(length)
19' 6" 21' 7" 21' 7"Live Area
COPY AREAheight)
8' 8" 9' 7" 9' 7"Live Area
TUP & BOTTOMBLANKING
101/2"Top & Bottom
(21" total)
5"Top & Bottom(10" total)
'5"Top & Bottom
(10" total)
ENDBLANKING
19"(38" total)
61/2"(13" total)
61/2"(13" total)
FRAMEWIDTH
11" 11" 11"
Kind Exposed Area
24-sheet 7. x 8.
30-sheet 10. x
Bleed 13. x 14.
Bulletins may also be in bleed form, depending upon the effect you wish to
create. In addition, dramatic effects can be attained by extending certain elements
of the design beyond the ed6... of the bulletin. The limits on any extension are on
top: 5'6" extra; on each side and bottom: 2' extra each. From the dimensions given
(14' by 48'), the total area of a bulletin is 16.. A further restric-
tion on extensions is that the combined surface area of all space extensions must
6
OUTDOOfl ADVERTISING
not exceed 200 square feet. This would give a possible total area of 17.
An example of an embellished bulletin would be:
Try your skill at identifying the following as either posters or bulletins.
Find A) ratio, B) poster or bulletin,(check ratio)
18.
Advertiser:Agency:Art Directors:Copywriter:
A.
B.
C.
C) if poster: is it 24-, 30-sheet, or
bleed
if bulletin: is it embellished?
19.
YellowPages
"01)11!
400:,....
Distim, Inc.Ingalls Assoc. inc., Ad.Jim Bennette/Karl CastlemanCorso Danati
7
A.
B.
C.
Advertiser:Agency:Art Director:Copywriter:
Mich. Bell TelephoneRoss Roy, Inc.Robert LawsonDon 3toll
20.
OUTDOOR ADVERTISING
21.
Advertiser: Campbell Soup Co. Advertiser: Minneapolis Gas Co.Agency: BBDO Agency: Knox Reeves AdvertisingArt Direcor: Gene Bove Art Director: Tom DonovanCopywriter: Mary Ellen Campbell Copywriter: Mike Vukodinovich
A.
B.
C.
Revsonvs.Donohue 23
The California 500 at Ontario Sept S.
Advertiser:Agency:Art Director:Copywriter:
A.
B.
C.
Ontario Motor SpeedwayChiat/Day Inc.Hy YablonkaJay Chiat
A.
B.
C.
EN YOU'RE HAVING MORE THAN ONE
Advertiser:Agency:Art Director:
A.
B.
C.
Schaefer BeerBBDOBill Petti
Bill began to put his project together. He had discovered a lot about outdoor ad-
vertising that he didn't know before and it had been interesting researching his topic.
Now all he had to do was prepare his project for submission to Mr. Meier. The fun
was over. The work was about to begin.
8
SPACE
Bruce looked out of the window as he wondered, "Who cares about all this stuff
anyway? None of it's got anything to do with what I'm going to do."
It was obvious that Mr. Nelson thought otherwise about geometry, the way he was
handling the unit. However, Bruces' thoughts were someplace else. He couldn't
wait to get home and assemble the six-inch newtonian reflecting telescope that
he had bought 77esterday. He had done all sorts of odd jobs to raise the money
and managed to con his dad into kicking in 20 percent of the cost. Actually,
his dad had said that he couldn't raise the money and he had bet him tle 20
percent that he could. They agreed to give his mom a different story because
she frowned on gambling in any form. Anyway, Bruce had the telescope and was
eager to get it in working order for tonight. He hoped to work directly in space
exploration later if he could ever manage to get out of high school.
"Bruce, have you figured it out yet?" Mr. Nelson's voice shattered his
train of thought. "Which project are you planning to work on?"
None of the available topics really excited him too much, so he asked Mr. Nelson
if he could talk to him after class and decide then. Mr. Nelson was busy,
but suggested a later hour. Since Bruce didn't have any classes then, they met
in the classroom.
They had quite a discussion on the relevancy of geometry, especially as it related
to Bruce's ambitions. The result of it was that Bruce's project for geometry
was to show its uses in astronomy and space-related situations and co write up
a report on a career in this area. "For once," thought Bruce, "I'm going to work
on something I like."
It seemed reasonable to start by finding a few facts about an astronomer's job.
SPACE
Bruce wrote to the American Astronomical Society in New Jersey and got back some
pamphlets. He discovered that astronomy is divided into many specialized branches.
Bruce liked me astronomy which involves the "measurement of angular positions
and movements of celestial bodies." Astrophysics, radio astronomy, and photometry
were some of the other branches.
Bruce found that there aren't too many astronomers around and many of them have
continued their schooling to very advanced degrees six to eight years beyond
high school. He felt that if the courses he had to take were in the area or
astronomy, even all that schooling wouldn't be bad. He did find that much of
the schooling involved science and math related courses in addition to the
elective classes.
The future for astronomers was interesting to Bruce. He found that since they
are involved in the locations and orbits of planets, and they study the size and
shape of the earth and its atmosphere and make calculations affecting navigation
and measurement of time they will play a very important role. The whole area
of space and programs involving satellites and exploration of space will also
greatly affect this career area.
Bruce felt that with all the schooling required and the responsibilities, certainly
he could expect to earn great money in that work: The average salary for those with
a PHD degree, in 1970, was $15,000. This satisfied Bruce and he decied to start
"zeroing in" on the geometry involved in space careers.
Bruce next found that problems range from the pure and oviously geometrical
problems to some which seemed at first glance only vaguely related to geometry.
In the first categor7 would be problems concerning solar cells, area and load
2
SPACE
on tbe feet of a moon lanching craft, and the distance to the horizon from a
given altitude above earth or the moon. Other problems based on geometry but
also requiring other branches of mathem:Ltics for their solution would include
transforming a rectangular map into an isosceles trapezoidal map, determining
relationships of volumes and areas in spacecraft pressure and storage tanks,
measuring the distance between earth and Mars. Following is a representative
selection of what Bruce found.
1. Solar cells convert the energy of sunlight directly into electrical energy.
For each square centimeter of solar cell in direct overhead sunlight, about 0.01
watt of electrical power is available. A solar cell in the shape of a regular
hexagon is required to deliver 10.4 watts. Find the minimum length of a side.
Procedure
The total area required is 10.4 watts
; .01 watt per square centimeter,
or a. square centimeters. The
regular hexagon can be partitioned in-
to six congruent triangles, each with
an area of b. square centimeters.
2The area of an equilateral triangle with side s is: A =
Solving this would give s = c.
2. Solar cells are made in various shapes to use most of the lateral
area of satellites. A certain circular solar cell with radius r
will produce 5 watts. Two equivalent solar cells are made, one being
a square with side s and the other an equilateral triangle with side
p. Find r in terms of p and also in terms of s.
3
SPACE
Procedure to find r in terms of s:
A (circle) = A (square)
fr r 2 = s2
Answer: r = a.
To find r in terms of p:
A (circle) = A (equilateral triangle).
Answer: r = b.
3. A spacecraft is at P, at an altitude h above earth's surface, as
pictured in the accompanying drawing. The distance to the horizon
is d, and r is the radius of the earth.
a. Derive an equation for d in terms of r and h.
Procedure
PA is tangent to the circle at A;
so j PAO is a right angle
r2 + d2 = (r+h) 2etc.
Answer:d =
b. Find the distance to the horizon if h =, 100 miles. Use 3,960
miles for the radius of earth.
Answer: d =
c. It is apparent that for near-earth orbits, h will to small
in comparison with r, so that discarding the h2 term intro-
duces only a mall error. The formula then simplifies to
d =4777 Find d with the simplified formula, and compute
the percent of error that results when the h2 term is dropped.
4
SPACE
Procedure
Use d = 4777 and solve as in b above.
Answered =
answer of b answer of c
To find percent of error: d.
so that
e. = 5.6
5.6 = f. percent.
answer of b
4. Solve problem three with respect to the moon's horizon for a spacecraft
70 miles above the surface of the moon. Use 1,080 miles for the
radius of the moon.
Procedure
Find true distance as in part b of problem three.
Answer: d = a.
Find distance by simplified formula (as in part c)
Answer: d = b.
Find percent of error as in part c above.
Anwer: c.
5. The average angle subtended by the moon for an observer on earth is
0.520 or 0,00907 radian. If the average distance from an observer
on earth to the center of the moon is known to be 384,4.0 kilometers,
find the diameter of the moon. Assume POM is a right triangle.
5
Moon
SPACE
Procedure
Use the tangent function to find
the radius of the moon, and then the
dia1neter follows.
a.
Hint: Tan 0.26° = PMOM
Since Tan 0.26° = .00452,PM may be determined.
6. From earth, the planet Mercury appears to oscillate about the sun,
appearing at elongation (its maximum angular distance from the sun as
seen from earth) every 58 days. Earth and Mercury revolve in the
same direction, counterclockwise as viewed from the north pole of
the sun. Determine the period of revolution of the planet Mercury.
Procedure.
An elongation occurs as SE1M1.
Fifty-eight days later, Mercury is at elongation
on the other side of the sun and another 58 days
late). it is at the elongation SE2M2. During the
116 days, Mercury has traversed one revolution
plus the arc M1M2. Earth has traversed in 116
days the angular distance 116 x 2fle = 114°365
Now, the triangles SE1M1 and SE2M2 are
congruent. Arc E1E2 = 114° and therefore arc
M1M2 = 114°. Hence, the period of Mercury is:
360° x116 days days.360° + 114°
6
SPACE
7. Using the accompanying figure for these problems, take one radius of earth
AE to be 3,960 miles. A formula has been derived for finding what fraction
of the surface of a sphere of radius r can be seen from an altitudQ h auovL
the surface of sphere. Let Az be the area of the zone with altitude BE,
Aebe the area of the earth, then
Az h
Ae 7Ti-1417
Gemini 10, with astronauts
Collins ,and Young aborad,
in an orbit with perigee of 100
miles and apogee of 168 miles.
What percent of earth's surface
was visible from each of these
two altitudes?
Procedure
Let h = 100; r = 3960, then Az =2(r+h)
Answer: a.
Let h = 168; r = 3960, then Az =W
Answer: b.
C. Find what altitude from earth the astronaut must be to see one-quarter of
earth's surface at one time.
procedure
Let Az . 1 .AT 4 2(3960 + h) etc.
Answer: h = c.
7
The first astronauts to travel that far from eatth were Anders,
Borman, and Lovell, on board Apollo 8, which orbited the moon m
Christmas Day, 1968.
d. What percent of ,,larth's surface cast be "seen" from a synchronous
satellite, whose altitude is 22,300 miles above earth?
Procedure
AzLet h = 22,300, A.;.=
Answer
A synchronous satellite can relay messages to about e.
percent of the earth's surface. Thus thr:e such satellites, evenly
spaced around the earth over the equator, could form the basis of
a communications network covering the entire earth
8
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